Past JEE Main Entrance Paper

Hint

(d)

$\begin{aligned} & \frac{3^{200}}{8}=\frac{1}{8}\left(9^{100}\right) \\ & =\frac{1}{8}(1+8)^{100}=\frac{1}{8}\left[1+n \cdot 8+\frac{n(n+1)}{2} \cdot 8^2+\ldots\right] \\ & =\frac{1}{8}+\text { Integer } \\ & \therefore\left\{\frac{3^{200}}{8}\right\}=\left\{\frac{1}{8}+\text { integer }\right\}=\frac{1}{8} \end{aligned}$

Hint

$(x+a)^n+(x-a)^n=2\left(T_1+T_3+T_5+\ldots . .\right)$
$\begin{aligned} & \left(\mathrm{x}+\sqrt{\mathrm{x}^2-1}\right)^6+\left(\mathrm{x}-\sqrt{\mathrm{x}^2-1}\right)^6 \\ & =2\left[T_1+T_3+T_5+T_7\right] \\ & =2\left[{ }^6 \mathrm{C}_0 \mathrm{x}^6+{ }^6 \mathrm{C}_2 \mathrm{x}^4\left(\mathrm{x}^2-1\right)+{ }^6 \mathrm{C}_4 \mathrm{x}^2\left(\mathrm{x}^2-1\right)^2+{ }^6 \mathrm{C}_6\left(\mathrm{x}^2-1\right)^3\right] \\ & =2\left[\mathrm{x}^6+15\left(\mathrm{x}^6-\mathrm{x}^4\right)+15 \mathrm{x}^2\left(\mathrm{x}^4-2 \mathrm{x}^2+1\right)+\left(-1+3 \mathrm{x}^2-3 \mathrm{x}^2+\mathrm{x}^6\right)\right] \\ & =2\left[32 \mathrm{x}^6-48 \mathrm{x}^4+18 \mathrm{x}^2-1\right] \\ & =64 \mathrm{x}^6-96 \mathrm{x}^4+36 \mathrm{x}^2-2 \\ & \alpha=-96 \\ & \beta=36 \\ & \therefore \alpha-\beta=-96-36 \\ & \quad=-132 \end{aligned}$

Hint

Given,
$\left(x^2+\frac{1}{x^3}\right)^n$, its $(r+1)^{\text {th }}$ term, is
$\begin{aligned} T_{r+1} & ={ }^n C_r x^{2 n-2 r} \cdot x^{-3 r} \\ & ={ }^n C_r x^{2 n-5 r} \end{aligned}$
$\begin{aligned} & \therefore 2 n-5 r=1 \\ & \Rightarrow r=\frac{2 n-1}{5} \end{aligned}$
$\begin{aligned} & \therefore \text { Coeff. of } x={ }^n C_{\left(\frac{2 n-1}{5}\right)}={ }^n C_{23} \\ & \therefore \frac{2 n-1}{5}=23 \text { or } n-\left(\frac{2 n-1}{5}\right)=23 \\ & \Rightarrow n=58 \text { or } n=38 \end{aligned}$
Minimum value is $n=38$

Hint

$\begin{aligned} & \left(\frac{2}{x}+x^{\log _8 x}\right)^6(x>0) \\ & \Rightarrow T_4=20 \times 8^7 \\ & \Rightarrow{ }^6 C_3\left(\frac{2}{x}\right)^3\left(x^{\log _8 x}\right)^3=20 \times 8^7 \\ & \Rightarrow \frac{160}{x^3} x^{3 \log _8 x}=20 \times 8^7 \\ & \Rightarrow x^{3 \log _8 x-3}=8^6 \\ & \Rightarrow x^{\log _2 x-3}=8^6=2^{18} \\ & \Rightarrow \log _2\left(x^{\log _2 x-3}\right)=\log _2 2^{18} \\ & \Rightarrow\left(\log _2 x-3\right)\left(\log _2 x\right)=18 \\ & \text { Let } \log _2 x =t \\ & \Rightarrow t^2-3 t-18=0 \\ & \Rightarrow t=6,-3 \\ & \Rightarrow \log _2 x=6 \quad \Rightarrow x=2^6=8^2 \\ & \Rightarrow \log _2 x=-3 \quad \Rightarrow x=2^{-3}=1 / 8 \\ \end{aligned}$

Hint

$\begin{aligned} & \left(\mathrm{x}+\sqrt{\mathrm{x}^3-1}\right)^6+\left(\mathrm{x}-\sqrt{\mathrm{x}^3-1}\right)^6 \\ & =2\left[{ }^6 \mathrm{C}_0 \mathrm{x}^6+{ }^6 \mathrm{C}_2 \mathrm{x}^4\left(\mathrm{x}^3-1\right)+{ }^6 \mathrm{C}_4 \mathrm{x}^2\left(\mathrm{x}^3-1\right)^2+{ }^6 \mathrm{C}_6\left(\mathrm{x}^3-1\right)^3\right] \\ & =2\left[{ }^6 \mathrm{C}_0 \mathrm{x}^6+{ }^6 \mathrm{C}_2 \mathrm{x}^7 – { }^6 \mathrm{C}_2 \mathrm{x}^4+{ }^6 \mathrm{C}_4 \mathrm{x}^8+{ }^6 \mathrm{C}_4 \mathrm{x}^2-2^6 \mathrm{C}_4 \mathrm{x}^5+\left(\mathrm{x}^9-1-3 \mathrm{x}^6+3 \mathrm{x}^3\right)\right] \\ & \Rightarrow \text { sum of coefficient of even powers of } \mathrm{x}=2[1-15+15+15-1-3]=24 \end{aligned}$

Hint

$\begin{aligned} & (x+10)^{50}={ }^{50} C_0(10)^{50}+{ }^{50} C_1 x \cdot 10^{49}+{ }^{50} C_2 x^2 \cdot 10^{48}+\cdots(1) \\ & (x-10)^{50}={ }^{50} C_0(10)^{50}-{ }^{50} C_1 x \cdot 10^{49}+{ }^{50} C_2 x^2 \cdot 10^{48}-\cdots(2) \end{aligned}$
Add equation (1) and (2)
$\begin{aligned} & (x+10)^{50}+(x-10)^{50} \\ & =2 \cdot{ }^{50} C_0 \cdot(10)^{50}+2 \cdot{ }^{50} C_2(10)^{48} \cdot x^2+\cdots \\ & \therefore a_0=2 \cdot{ }^{50} C_0(10)^{50}, a_2=2 \cdot{ }^{50} C_2(10)^{48} \\ & \Rightarrow \frac{a_2}{a_0}=\frac{2 \cdot{ }^{50} C_2(10)^{48}}{2 \cdot{ }^{50} C_0(10)^{50}} \\ & \therefore \frac{a_2}{a_0}=\frac{50 \times 49}{2 \times 100}=12.25 \end{aligned}$

Hint

(a) Third term of $\left(1+x^{\log _2 x}\right)^5={ }^5 C_2\left(x^{\log _2 x}\right)^{5-3}$
$={ }^5 C_2\left(x^{\log _2 x}\right)^2$
Given, ${ }^5 \mathrm{C}_2\left(x^{\log _2 x}\right)^2=2560$
$\begin{aligned} & \Rightarrow \quad\left(x^{\log _2 x}\right)^2=256=(\pm 16)^2 \\ & \Rightarrow \quad x^{\log _2 x}=16 \text { or } x^{\log _2 x=-16 \text { (rejected) }} \\ & \Rightarrow \quad x^{\log _2 x}=16 \Rightarrow \log _2 x \log _2 x=\log _2 16=4 \\ & \Rightarrow \quad \log _2 x=\pm 2 \Rightarrow x=2^2 \text { or } 2^{-2} \\ & \Rightarrow x=4 \text { or } \frac{1}{4} \end{aligned}$

Hint

$\begin{aligned} & \frac{2^{403}}{15}=2^3 \times\left(\frac{2^{400}}{15}\right)=8 \times\left(\frac{16^{100}}{15}\right) \\ & =\frac{8}{15}(1+15)^{100} \end{aligned}$
Using binomial theorem we can write above expression as
$\begin{aligned} & \frac{8}{15}(1+15 . n), n \in N \\ & =\frac{8}{15}+8 . n \end{aligned}$
So the fractional part is $\frac{8}{15}=\frac{k}{15}$
Hence, the value of $k=8$

Hint

(a) Let $a=\left(\left(1+2 x+3 x^2\right)^6+\left(1-4 x^2\right)^6\right)$
$\therefore$ Coefficient of $x^2$ in the expansion of the product $\left(2-x^2\right)\left(\left(1+2 x+3 x^2\right)^6+\left(1-4 x^2\right)^6\right)$ $=2$ (Coefficient of $x^2$ in a) $-1$ (Constant of expansion) In the expansion of $\left(\left(1+2 x+3 x^2\right)^6+\left(1-4 x^2\right)^6\right)$.
Constant $=1+1=2$
Coefficient of $x^2=\left[\right.$ Coefficient of $x^2$ in $\left.\left({ }^6 C_0(1+2 x)^6\left(3 x^2\right)^0\right)\right]+[$ Coefficient of $x^2$ in $\left.\left({ }^6 C_1(1+2 x)^5\left(3 x^2\right)^1\right)\right]$
– Coefficient of $x^2$ in $\left[{ }^6 \mathrm{C}_1\left(4 x^2\right)\right]$
$=60+6 \times 3-24=54$
$\therefore$ The coefficient of $x^2$ in
$\begin{aligned} & \left(2-x^2\right)\left(\left(1+2 x+3 x^2\right)^6+\right. \\ & \left.\left(1-4 x^2\right)^6\right) \\ & =2 \times 54-1(2)=108-2=106 \end{aligned}$

Hint

(c) Since we know that,
$\begin{aligned} & (\mathrm{x}+\mathrm{a})^5+(\mathrm{x}-\mathrm{a})^5 \\ & =2\left[{ }^5 \mathrm{C}_0 \mathrm{x}^5+{ }^5 \mathrm{C}_2 \mathrm{x}^3 \cdot \mathrm{a}^2+{ }^5 \mathrm{C}_4 \mathrm{x} \cdot \mathrm{a}^4\right] \\ & \quad \therefore \quad\left(\mathrm{x}+\sqrt{\mathrm{x}^3-1}\right)^5+\left(\mathrm{x}-\sqrt{\mathrm{x}^3-1}\right)^5 \\ & =2\left[{ }^5 \mathrm{C}_0 \mathrm{x}^5+{ }^5 \mathrm{C}_2 \mathrm{x}^3\left(\mathrm{x}^3-1\right)+{ }^5 \mathrm{C}_4 \mathrm{x}\left(\mathrm{x}^3-1\right)^2\right] \\ & \quad \Rightarrow \quad 2\left[\mathrm{x}^5+10 \mathrm{x}^6-10 \mathrm{x}^3+5 \mathrm{x}^7-10 \mathrm{x}^4+5 \mathrm{x}\right] \end{aligned}$
$\therefore$ Sum of coefficients of odd degree terms $=2$.

Hint

(a) $\left[\frac{\left(x^{1 / 3}+1\right)\left(x^{2 / 3}-x^{1 / 3}+1\right)}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}-\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}\right]^{10}$
$\begin{aligned} & =\left(\mathrm{x}^{1 / 3}+1-1-1 / \mathrm{x}^{1 / 2}\right)^{10}=\left(x^{1 / 3}-1 / x^{1 / 2}\right)^{10} \\ & \mathrm{~T}_{r+1}={ }^{10} C_{\mathrm{r}} \frac{20-5 \mathrm{r}}{6} \\ & \text { for } \mathrm{r}=10 \\ & \quad \mathrm{~T}_{11}={ }^{10} \mathrm{C}_{10} \mathrm{x}^{-5} \end{aligned}$
Coefficient of $\mathrm{x}^{-5}={ }^{10} C_{10}(1)(-1)^{10}=1$

Hint

$\begin{aligned} & \text { (b) } T_{r+1}={ }^{18} C_r\left(x^{\frac{1}{3}}\right)^{18-r}\left(\frac{1}{2 x^{\frac{1}{3}}}\right)^r={ }^{18} C_r x^{6-\frac{2 r}{3}} \frac{1}{2^r} \\ & \left\{\begin{array}{l} 6-\frac{2 r}{3}=-2 \Rightarrow r=12 \\ \& 6-\frac{2 r}{3}=-4 \Rightarrow r=15 \end{array}\right\} \end{aligned}$

$\Rightarrow \frac{\text { coefficient of } x^{-2}}{\text { coefficient of } x^{-4}}=\frac{{ }^{18} C_{12} \frac{1}{2^{12}}}{{ }^{18} C_{15} \frac{1}{2^{15}}}=182$

Hint

$\begin{aligned} & \left(1+a x+b x^2\right)(1-2 x)^{18}=1(1-2 x)^{18}+a x(1-2 x)^{18}+b x^2(1-2 x)^{18} \\ & \text { Coefficient of } x^3:(-2)^3{ }^{18} C_3+a(-2)^2{ }^{18} C_2+b(-2)^{18} C_1=0 \\ & \frac{4 \times(17 \times 16)}{(3 \times 2)}-2 a \times \frac{17}{2}+b=0-(1) \\ & \text { Coefficient of } x^4:(-2)^4{ }^{18} C_4+a(-2)^3{ }^{18} C_3+b(-2)^2{ }^{18} C_2=0 \\ & (4 \times 20)-2 a \times \frac{16}{3}+b=0-(2) \\ & \text { From equations }(1) \text { and }(2), \text { we get } \\ & 4\left(\frac{17 \times 8}{3}-20\right)+2 a\left(\frac{16}{3}-\frac{17}{2}\right)=0 \\ & \Rightarrow a=16 \\ & \Rightarrow b=\frac{2 \times 16 \times 16}{3}-80=\frac{272}{3} \end{aligned}$

Hint

$\begin{aligned} \because X & =\left\{4^n-3 n-1: n \in N\right\} \\ X & =\{0,9,54,243, \ldots\} \text { [put } n=1,2,3, \ldots] \\ Y & =\{9(n-1): n \in N\} \\ Y & =\{0,1,18,27, \ldots\} \end{aligned}$
It is clear that $X \subset Y$.
$\therefore \quad X \cup Y=Y$

Hint

(c) Given expansion is
$\begin{aligned} &(1+x)^{101}\left(1-x+x^2\right)^{100} \\ &=(1+x)(1+x)^{100}\left(1-x+x^2\right)^{100} \\ &=(1+x)\left[(1+x)\left(1-x+x^2\right)\right]^{100} \\ &=(1+x)\left[\left(1-x^3\right)^{100}\right] \end{aligned}$
Expansion $\left(1-x^3\right)^{100}$ will have $100+1=101$ terms.
So, $(1+x)\left(1-x^3\right)^{100}$ will have $2 \times 101=202$ terms

Hint

$\text { (d) }\left(2^{1 / 2}+3^{1 / 5}\right)^{10}={ }^{10} \mathrm{C}_0\left(2^{1 / 2}\right)^{10} +{ }^{10} \mathrm{C}_1\left(2^{1 / 2}\right)^9\left(3^{1 / 5}\right)+\ldots \ldots+{ }^{10} \mathrm{C}_{10}\left(3^{1 / 5}\right)^{10}$
There are only two rational terms – first term and last term. Now sum of two rational terms
$=(2)^5+(3)^2=32+9=41$

Hint

Let $y=\sum_{r=1}^{10} A_r\left(B_{10} B_r-C_{10} A_r\right)$.
Now, $\sum_{r=1}^{10} A_r B_r=$ coefficient of $x^{10}$ in $\left[(1+x)^{10}(1+x)^{20}\right]-1$
$\Rightarrow \sum_{\mathrm{r}=1}^{10} \mathrm{~A}_{\mathrm{r}} \mathrm{B}_{\mathrm{r}}=\mathrm{C}_{20}-1=\mathrm{C}_{10}-1$
And $\sum_{r=1}^{10} A_r^2=$ coefficient of $x^{10}$ in $\left[(1+x)^{10}(1+x)^{10}\right]-1$ $\Rightarrow \sum_{\mathrm{r}=1}^{10} \mathrm{~A}_{\mathrm{r}}^2=\mathrm{B}_{10}-1$

Therefore, $\mathrm{y}=\mathrm{B}_{10}\left(\mathrm{C}_{10}-1\right)-\mathrm{C}_{10}\left(\mathrm{~B}_{10}-1\right)=\mathrm{C}_{10}-\mathrm{B}_{10}$.

Hint

(d) $\left(1+t^2\right)^{12}\left(1+t^{12}\right)\left(1+t^{24}\right)$
$\begin{aligned} & =\left(1+t^{12}+t^{24}+t^{36}\right)\left(1+t^2\right)^{12} \\ & \therefore \text { Coeff. of } t^{24}=1 \times \text { Coeff. of } t^{24} \text { in }\left(1+t^2\right)^{12}+1 \times \\ & \text { Coeff. of } t^{12} \text { in }\left(1+t^2\right)^{12}+1 \times \text { constant term in }\left(1+t^2\right)^{12} \\ & ={ }^{12} C_{12}+{ }^{12} C_6+{ }^{12} C_0=1+{ }^{12} C_6+1={ }^{12} C_6+2 \\ & \end{aligned}$

Hint

(b) In binomial expansion $(a-b)^n, \mathrm{n} \geq 5$;
$\begin{aligned} & T_5+T_6=0 \\ & \Rightarrow{ }^n C_4 a^{n-4} b^4-{ }^n C_5  a^{n-5} b^5=0 \\ & \Rightarrow \frac{{ }^n C_4}{{ }^n C_5} \cdot \frac{a}{b}=1 \Rightarrow \frac{5}{n-4} \cdot \frac{a}{b}=1 \Rightarrow \quad \frac{a}{b}=\frac{n-4}{5} \end{aligned}$

Hint

(a) General term in the expansion $\left(\frac{x}{2}-\frac{3}{x^2}\right)^{10}$ is
$T_{r+1}={ }^{10} C_r\left(\frac{x}{2}\right)^{10-r}\left(\frac{-3}{x^2}\right)^r={ }^{10} C_r x^{10-3 r} \frac{(-1)^r 3^r}{2^{10-r}}$
To find coeff of $x^4$, put $10-3 \mathrm{r}=4 \Rightarrow \mathrm{r}=2$
$\therefore$ Coeff of $\mathrm{x}^4={ }^{10} \mathrm{C}_2 \frac{(-1)^2 3^2}{2^8}=\frac{405}{256}$

Hint

(a) Given : $\mathrm{r}$ and $\mathrm{n}$ are positive integers such that $\mathrm{r}>1, \mathrm{n}>2$
Also, in the expansion of $(1+x)^{2 n}$
Coeff. of $(3 \mathrm{r})^{\text {th }}$ term $=$ Coeff. of $(\mathrm{r}+2)^{\mathrm{th}}$ term
$\begin{aligned} & \Rightarrow{ }^{2 \mathrm{n}} \mathrm{C}_{3 \mathrm{r}-1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1} \\ & \Rightarrow 3 \mathrm{r}-1=\mathrm{r}+1 \text { or } 3 \mathrm{r}-1+\mathrm{r}+1=2 \mathrm{n} \\ & {\left[\because \text { If }{ }^n C_p={ }^n C_q \text {, then } p=q \text { or } p+q=n\right]} \\ & \Rightarrow \mathrm{r}=1 \text { or } 2 \mathrm{r}=\mathrm{n} \\ & \end{aligned}$
But $\mathrm{r}>1 \quad \therefore \quad \mathrm{n}=2 \mathrm{r}$

Hint

$\begin{aligned} & \sum_{\mathrm{r}=0}^{\mathrm{n}} \mathrm{r}\left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\right)^2=\mathrm{n} \sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1} \\ & =\mathrm{n} \sum_{\mathrm{r}=1}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}}{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}={ }^{2 \mathrm{n}-1} \mathrm{C}_{\mathrm{n}-1} \end{aligned}$
Now,
$\begin{aligned} \mathrm{X} & =\left({ }^{10} \mathrm{C}_1\right)^2+2\left({ }^{10} \mathrm{C}_2\right)^2+3\left({ }^{10} \mathrm{C}_3\right)^2+\ldots+10\left({ }^{10} \mathrm{C}_{10}\right)^2 \\ & =\sum_{\mathrm{n}=0}^{10} \mathrm{r}\left({ }^{10} \mathrm{C}_{\mathrm{r}}\right)^2=10^{19} \mathrm{C}_9 \\ \therefore & \frac{\mathrm{X}}{1430}=\frac{1}{143}{ }^{19} \mathrm{C}_9=646 \end{aligned}$

Hint

(5) $(1+x)^2+(1+x)^3+\ldots .+(1+x)^{49}+(1+m x)^{50}$
$\begin{aligned} & =(1+x)^2\left[\frac{(1+x)^{48}-1}{(1+x)-1}\right]+(1+m x)^{50} \\ & =\frac{1}{x}\left[(1+x)^{50}-(1+x)^2\right]+(1+m x)^{50} \end{aligned}$
Coeff. of $x^2$ in the above expansion
$\begin{aligned} & =\text { Coeff. of } x^3 \text { in }(1+x)^{50}+\text { Coeff. of } x^2 \text { in }(1+m x)^{50} \\ & ={ }^{50} \mathrm{C}_3+{ }^{50} \mathrm{C}_2 \mathrm{~m}^2 \\ & \therefore(3 \mathrm{n}+1){ }^{51} \mathrm{C}_3={ }^{50} \mathrm{C}_3+{ }^{50} \mathrm{C}_2 \mathrm{~m}^2 \\ & \Rightarrow(3 \mathrm{n}+1)=\frac{{ }^{50} \mathrm{C}_3}{{ }^{51} \mathrm{C}_3}+\frac{{ }^{50} \mathrm{C}_2}{{ }^{51} \mathrm{C}_3} \mathrm{~m}^2 \\ & \Rightarrow 3 \mathrm{n}+1=\frac{16}{17}+\frac{1}{17} \mathrm{~m}^2 \Rightarrow \mathrm{n}=\frac{\mathrm{m}^2-1}{51} \\ & \end{aligned}$
$\therefore$ Least positive integer $\mathrm{m}$ for which $\mathrm{n}$ is an integer is m $=16$ and then $n=5$

Hint

(6) Let the coefficients of three consecutive terms of $(1+x)^{n+5}$ be ${ }^{n+}$ ${ }^5 C_{r-1},{ }^{n+5} C_r,{ }^{n+5} C_{r+1}$, then we have
$\begin{aligned} & { }^{n+5} C_{r-1}:{ }^{n+5} C_r:{ }^{n+5} C_{r+1}=5: 10: 14 \\ & \frac{{ }^{n+5} C_{r-1}}{{ }^{n+5} C_r}=\frac{5}{10} \Rightarrow \frac{r}{n+6-r}=\frac{1}{2} \end{aligned}$
$\begin{aligned} & \Rightarrow n-3 r+6=0 \\ \text { Also } & \frac{{ }^{n+5} C_r}{{ }^{n+5} C_{r+1}}=\frac{10}{14} \Rightarrow \frac{r+1}{n-r+5}=\frac{5}{7} \\ & \Rightarrow 5 n-12 r+18=0 \end{aligned}$
Solving (i) and (ii), we get $n=6$.

Hint

(13)
$\begin{aligned} & T_{r+1}={ }^{22} C_r \cdot\left(x^m\right)^{22-r} \cdot\left(\frac{1}{x^2}\right)^r \\ & T_{r+1}={ }^{22} C_r \cdot x^{22 m-m r-2 r} \\ & \because 22 m-m r-2 r=1 \\ & \Rightarrow r=\frac{22 m-1}{m+2} \Rightarrow r=22-\frac{3 \cdot 3 \cdot 5}{m+2} \end{aligned}$
So, possible value of $m=1,3,7,13,43$
But ${ }^{22} C_r=1540$
$\therefore$ Only possible value of $m=13$.

Hint

$\begin{aligned} & \text { Coefficient of } x^4 \text { in }\left(\frac{1-x^4}{1-x}\right)^6=\text { coefficient of } x^4 \text { in }\left(1-6 x^4\right)(1-x)^{-6} \\ & =\text { coefficient of } x^4 \text { in }\left(1-6 x^4\right)\left[1+{ }^6 C_1 x+{ }^7 C_2 x^2+\ldots\right] \\ & \quad={ }^9 C_4-6 \cdot 1=126-6=120 \end{aligned}$

Hint

Finding the value of $\frac{a_7}{a_{13}}$ :
Given, $\left(2 x^2+3 x+4\right)^{10}=\sum_{r=0}^{20} a_r x^r \quad \ldots \ldots(1)$
To obtain the symmetry, replace $x$ by $\frac{2}{x}$ in above equation we get,
$\begin{aligned} & {\left[2\left(\frac{2}{x}\right)^2+3\left(\frac{2}{x}\right)+4\right]^{10}=\sum_{r=0}^{20} a_r \frac{2^r}{x^r}} \\ & \Rightarrow \quad\left[\frac{8}{x^2}+\frac{6}{x}+4\right]^{10}=\sum_{r=0}^{20} a_r \times 2^r\left(\frac{1}{x^r}\right) \\ & \Rightarrow \quad \frac{2^{10}}{x^{20}}\left[2 x^2+3 x+4\right]^{10}=\sum_{r=0}^{20} a_r 2^r\left(\frac{1}{x^r}\right) \\ & \Rightarrow \quad\left[2 x^2+3 x+4\right]^{10}=\sum_{r=0}^{20} a_r 2^{(r-10)} x^{(20-r)} \\ & \end{aligned}$
Substituting equation (1) in equation (2), we get
$\sum_{r=0}^{20} a_r x^r=\sum_{r=0}^{20} a_r 2^{(r-10)} x^{(20-r)}$
Put $r=7$ in LHS and $r=13$ on RHS to find the coefficients of $x^7$ on both sides and compare them we get,
$\begin{aligned} & a_7= a_{13}(2)^3 \\ & \Rightarrow \frac{a_7}{a_{13}}=8 \end{aligned}$
Hence, the value of the ratio $\frac{a_7}{a_{13}}=8$.

Hint

Given expression : $\left(\sqrt{2}+3^{1 / 5}\right)^{10}$
$\begin{aligned} \therefore & T_{r+1}={ }^{10} C_r(\sqrt{2})^{10-r} \cdot\left(3^{1 / 5}\right)^r \cdot(0 \leq r \leq 10) \\ & =\frac{10 !}{r !(10-r) !} \cdot 2^{5-r / 2} \cdot 3^{r / 5} \end{aligned}$
$\mathrm{T}_{\mathrm{r}+1}$ will be rational if $2^{5-r / 2}$ and $3^{\mathrm{r} / 5}$ are rational numbers.
$\Rightarrow 5-\frac{r}{2}$ and $\frac{r}{5}$ are integers
$\Rightarrow \mathrm{r}=0$ and $\mathrm{r}=10 \Rightarrow \mathrm{T}_1$ and $\mathrm{T}_{11}$ are rational terms.
Now, $\mathrm{T}_1+\mathrm{T}_{11}={ }^{10} \mathrm{C}_0 2^{5-0} \cdot 3^0+{ }^{10} \mathrm{C}_{10} 2^{5-5} \cdot 3^2$
$=1.32 .1+1.1 .9=32+9=41$

Hint

We know that for $a$ positive integer $n$
$(1+x)^n={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+\ldots . .+{ }^n C_n x^n$
Since coefficients of $2^{\text {nd }}, 3^{\text {rd }}$ and $4^{\text {th }}$ terms are in A.P.
$\begin{aligned} & \therefore{ }^n C_1,{ }^n C_2,{ }^n C_3 \text { are in A.P. } \\ & \Rightarrow 2 \cdot{ }^n C_2={ }^n C_1+{ }^n C_3 \\ & \Rightarrow 2 \times \frac{n(n-1)}{2}=n+\frac{n(n-1)(n-2)}{3 !} \\ & \Rightarrow n-1=1+\frac{n^2-3 n+2}{6} \Rightarrow n^2-9 n+14=0 \\ & \Rightarrow(n-7)(n-2)=0 \Rightarrow \mathrm{n}=7 \text { or } 2 \end{aligned}$
But for the existence of $4^{\text {th }}$ term, $n=7$.

Hint

$\begin{aligned} & (101)^{50}-\left\{(99)^{50}+(100)^{50}\right\} \\ & =(100+1)^{50}-(100-1)^{50}-(100)^{50} \\ & =(100)^{50}\left[(1+0.01)^{50}-(1-0.01)^{50}-1\right] \\ & =(100)^{50}\left[2\left(\left(^{50} C_1(0.01)+{ }^{50} C_3(0.01)^3+\ldots .\right)-1\right]\right. \\ & \quad=(100)^{50}\left[2 \times 50 \times \frac{1}{100}+2\left({ }^{50} C_3(0.01)^3+\ldots .\right)-1\right] \\ & =(100)^{50}\left[2\left(\left(^{50} C_3(0.01)^3+\ldots .\right)\right]>0\right. \\ & \quad \therefore(101)^{50}>(99)^{50}+(100)^{50} \quad \therefore(101)^{50} \text { is greater. } \end{aligned}$

Hint

$\begin{aligned} & \text { (c) Let } b=\sum_{r=0}^n \frac{r}{{ }^n C_r}=\sum_{r=0}^n \frac{n-(n-r)}{{ }^n C_r} \\ & =n a_n-\sum_{r=0}^n \frac{n-r}{{ }^n C_{n-r}} \quad\left[\because{ }^n C_r={ }^n C_{n-r}\right] \\ & =n a_n-b \\ & \Rightarrow 2 b=n a_n \Rightarrow b=\frac{n}{2} a_n \\ & \end{aligned}$

Hint

$\text { (c) } \begin{aligned} \text { General term }=T_{r+1}= & { }^{10} C_r(\sqrt{x})^{10-r} \cdot\left(-\frac{k}{x^2}\right)^r \\ & ={ }^{10} C_r(-k)^r \cdot x^{\frac{10-r}{2}-2 r} \\ & ={ }^{10} C_r(-k)^r \cdot x^{\frac{10-5 r}{2}} \end{aligned}$
Since, it is constant term, then
$\begin{aligned} & \frac{10-5 r}{2}=0 \Rightarrow r=2 \\ & \therefore{ }^{10} C_2(-k)^2=405 \\ & \Rightarrow k^2=\frac{405 \times 2}{10 \times 9}=\frac{81}{9}=9 \\ & \therefore|k|=3 \end{aligned}$

Hint

Let the three consecutive terms in the binomial expansion of $(1+\mathrm{x})^{\mathrm{n}+5}$ are ${ }^{\mathrm{n}+5} \mathrm{C}_{\mathrm{r}-1},{ }^{\mathrm{n}+5} \mathrm{C}_{\mathrm{r}}$ and ${ }^{\mathrm{n}+5} \mathrm{C}_{\mathrm{r}+1}$
Now, according to the given information ${ }^{\mathrm{n}+5} \mathrm{C}_{\mathrm{r}-1}:{ }^{\mathrm{n}+5} \mathrm{C}_{\mathrm{r}}:{ }^{\mathrm{n}+5} \mathrm{C}_{\mathrm{r}+1}=5: 10: 14$
$\Rightarrow \quad \frac{(\mathrm{n}+5) !}{(\mathrm{r}-1) !(\mathrm{n}-\mathrm{r}+6) !}: \frac{(\mathrm{n}+5) !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}+5) !}: \frac{(\mathrm{n}+5) !}{(\mathrm{r}+1) !(\mathrm{n}-\mathrm{r}+4) !}=5: 10: 14$
$\begin{aligned} & \Rightarrow \quad \frac{1}{(n-\mathrm{r}+6)(\mathrm{n}-\mathrm{r}+5)}: \frac{1}{\mathrm{r}(\mathrm{n}-\mathrm{r}+5)}: \frac{1}{(\mathrm{r}+1) \mathrm{r}} \\ & \text { So, } \quad \frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+6}=\frac{5}{10} \\ & \Rightarrow \quad 2 \mathrm{r}=\mathrm{n}-\mathrm{r}+6 \Rightarrow \mathrm{n}+6=3 \mathrm{r} \dots(i)\\ & \text { and } \quad \frac{\mathrm{r}+1}{\mathrm{n}-\mathrm{r}+5}=\frac{5}{7} \\ & \Rightarrow \quad 7 \mathrm{r}+7 \quad=5 \mathrm{n}-5 \mathrm{r}+25 \\ & \Rightarrow \quad 5 \mathrm{n}+18=12 \mathrm{r} \dots(ii) \end{aligned}$
From Eqs. (i) and (ii), we have $n=6$.
So, the largest coefficient in the expansion is same as the greatest binomial coefficient
$\begin{aligned} & ={ }^{11} \mathrm{C}_5 \text { or }{ }^{11} \mathrm{C}_6 \\ & =\frac{11 !}{5 ! 6 !}=\frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2}=462 \end{aligned}$

Hint

(c) Here, $\left(3^{\frac{1}{2}}+5^{\frac{1}{8}}\right)^n$
$T_{r+1}={ }^n C_r(3)^{\frac{n-r}{2}}(5)^{\frac{r}{8}}$
$\because \frac{n-r}{2}$ and $\frac{r}{8}$ are integer
So, $r$ must be $0,8,16,24 \ldots \ldots$
Now $n=t_{33}=a+(n-1) d=0+32 \times 8=256$
$\Rightarrow n=256$

Hint

(c) General term $=T_{r+1}={ }^9 C_r\left(\frac{3 x^2}{2}\right)^{9-r}\left(-\frac{1}{3 x}\right)^r$
$={ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{18-3 r}$
The term is independent of $x$, then
$\begin{aligned} & 18-3 r=0 \Rightarrow r=6 \\ & \therefore T_7={ }^9 C_6\left(\frac{3}{2}\right)^3\left(-\frac{1}{3}\right)^6={ }^9 C_3\left(\frac{1}{6}\right)^3 \\ & =\frac{9 \times 8 \times 7}{3 \times 2 \times 1}\left(\frac{1}{6}\right)^3=\left(\frac{7}{18}\right) . \\ & \therefore 18 k=18 \times \frac{7}{18}=7 . \end{aligned}$

Hint

(a) General term of
$\begin{gathered} \left(\alpha x^{\frac{1}{9}}+\beta x^{\frac{-1}{6}}\right)^{10}={ }^{10} C_r\left(\alpha x^{\frac{1}{9}}\right)^{10-r}\left(\beta x^{\frac{-1}{6}}\right)^r \\ ={ }^{10} C_r \alpha^{10-r} \beta^r(x)^{\frac{10-r}{9}-\frac{r}{6}} \end{gathered}$
Term independent of $x$ if $\frac{10-r}{9}-\frac{r}{6}=0 \Rightarrow r=4$.
$\therefore$ Term independent of $x={ }^{10} C_4 \alpha^6 \beta^4$
Since $\alpha^3+\beta^2=4$
Then, by AM-GM inequality
$\begin{aligned} & \frac{\alpha^3+\beta^2}{2} \geq\left(\alpha^3 b^2\right)^{\frac{1}{2}} \\ & \Rightarrow(2)^2 \geq \alpha^3 \beta^2 \Rightarrow \alpha^6 \beta^4 \leq 16 \end{aligned}$
$\because$ The maximum value of the term independent of $x=10 \mathrm{k}$
$\therefore 10 k={ }^{10} C_4 \cdot 16 \Rightarrow k=336 .$

Hint

(b) General term of the given expansion
$T_{r+1}={ }^{16} C_r\left(\frac{x}{\sin \theta}\right)^{16-r}\left(\frac{1}{x \cos \theta}\right)^r$
For $r=8$ term is free from ‘ $x$ ‘
$\begin{aligned} & T_9={ }^{16} C_8 \frac{1}{\sin ^8 \theta \cos ^8 \theta} \\ & T_9={ }^{16} C_8 \frac{2^8}{(\sin 2 \theta)^8} \end{aligned}$
When $\theta \in\left[\frac{\pi}{8}, \frac{\pi}{4}\right]$, then least value of the term independent of $x$,
$l_1={ }^{16} C_8 2^8$
[Q min. value of $l_1$ at $\theta=\pi / 4$ ]
When $\theta \in\left[\frac{\pi}{16}, \frac{\pi}{8}\right]$, then least value of the term independent of $x$,
$l_2={ }^{16} C_8=\frac{2^8}{\left(\frac{1}{\sqrt{2}}\right)^8}={ }^{16} \mathrm{C}_8 \cdot 2^8 \cdot 2^4$
[Q min. value of $l_2$ at $\theta=\pi / 8$ ]
Now, $\frac{l_2}{l_1}=\frac{{ }^{16} C_8 \cdot 2^8 \cdot 2^4}{{ }^{16} C_8 \cdot 2^8}=16: 1$

Hint

$\text { (d) Let the general term of the expansion }$

$\begin{gathered} T_{r+1}={ }^{60} C_r\left(7^{\frac{1}{5}}\right)^{60-r}\left(-3^{\frac{1}{10}}\right)^r \\ ={ }^{60} C_r \cdot(7)^{12-\frac{r}{5}}(-1)^r \cdot(3)^{\frac{r}{10}} \end{gathered}$
Then, for getting rational terms, $r$ should be multiple of L.C.M. of $(5,10)$ Then, $r$ can be $0,10,20,30,40,50,60$.
Since, total number of terms $=61$
Hence, total irrational terms $=61-7=54$

Hint

(c) $\left(2^{\frac{1}{3}}+\frac{1}{2(3)^{\frac{1}{3}}}\right)^{10}={ }^{10} C_0\left(2^{\frac{1}{3}}\right)^0\left(\frac{1}{2(3)^{1 / 3}}\right)^{10}+$
$\cdots+{ }^{10} C_{10}\left(2^{\frac{1}{3}}\right)^{10}\left(\frac{1}{2(3)^{1 / 3}}\right)^0$
$\begin{aligned} & 5^{\text {th }} \text { term from beginning } T_5={ }^{10} C_4\left(2^{\frac{1}{3}}\right)^6 \frac{1}{\left(2.3^{\frac{1}{3}}\right)^4}+{ }^{10} C_{10}\left(2^{\frac{1}{3}}\right)^{10}\left(\frac{1}{2(3)^{1 / 3}}\right)^0 \\ & \text { and } 5^{\text {th }} \text { term from end } T_{11-5+1}={ }^{10} C_6\left(2^{\frac{1}{3}}\right)^4\left(\frac{1}{2.3^{\frac{1}{3}}}\right)^6 \\ & \therefore T_5: T_7={ }^{10} C_4\left(2^{\frac{1}{3}}\right)^6\left(\frac{1}{2.3^{\frac{1}{3}}}\right)^4:{ }^{10} C_6\left(2^{\frac{1}{3}}\right)^4\left(\frac{1}{2.3^{\frac{1}{3}}}\right)^6 \\ & =\left(2^{\frac{1}{3}}\right)^2:\left(\frac{1}{2.3^{\frac{1}{3}}}\right)^2 \end{aligned}$
$=\frac{2^{\frac{2}{3}} \cdot 2^2 \cdot 3^{\frac{2}{3}}}{1}=4(6)^{\frac{2}{3}}: 1=4 \cdot(36)^{\frac{1}{3}}: 1$

Hint

(c) Given expression can be written as
$\begin{aligned} & \quad\left[\frac{\left(x^{1 / 3}\right)^3+1^3}{x^{2 / 3}-x^{1 / 3}+1}-\frac{(\sqrt{x})^2-1^2}{\sqrt{x}(\sqrt{x}-1)}\right]^{10} \\ & =\left(\left(x^{1 / 3}+1\right)-\left(\frac{\sqrt{x}+1}{\sqrt{x}}\right)\right)^{10}=\left(x^{1 / 3}+1-1-\frac{1}{\sqrt{x}}\right)^{10} \\ & =\left(x^{1 / 3}-x^{-1 / 2}\right)^{10} \\ & \text { General term }=\mathrm{T}_{r+1} \\ & ={ }^{10} \mathrm{C}_{\mathrm{r}}\left(x^{1 / 3}\right)^{10-r}\left(-x^{-1 / 2}\right)^r={ }^{10} C_r x^{\frac{10-r}{3}} \cdot(-1)^r \cdot x^{-\frac{r}{2}} \\ & ={ }^{10} C_r(-1)^r \cdot x^{\frac{10-r}{3}-\frac{r}{2}} \end{aligned}$
Term will be independent of $x$ when $\frac{10-r}{3}-\frac{r}{2}=0$
$\Rightarrow r=4$
So, required term $=\mathrm{T}_5={ }^{10} \mathrm{C}_4=210$

Hint

According to the question,
$\begin{aligned} & { }^n C_{r-1}:{ }^n C_r:{ }^n C_{r+1}=2: 5: 12 \\ & \Rightarrow \frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{5}{2} \Rightarrow \frac{n-r+1}{r}=\frac{5}{2} \\ & \Rightarrow 2 n-7 r+2=0 \dots(i)\\ & \frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{12}{5} \Rightarrow \frac{n-r}{r+1}=\frac{12}{5} \\ & \Rightarrow 5 n-17 r-12=0 \dots(ii) \end{aligned}$
Solving eqns. (i) and (ii),
$n=118, r=34$

Hint

Given, $\mathrm{n}$ is an even positive integer.
Let $n=2 m, \therefore k=3 m, n \in N$
$\begin{aligned} & \text { LHS } \sum_{r=1}^k(-3)^{r-1} \cdot{ }^{3 n} C_{2 r-1}=\sum_{r=1}^{3 m}(-3)^{r-1} \quad{ }^{6 m} C_{2 r-1} \\ & ={ }^{6 m} C_1-3 \cdot{ }^{6 m} C_3+3^2 \cdot{ }^{6 m} C_5 \\ & -\ldots+(-3)^{3 m-16 m} C_{6 m-1} \ldots \text { (i) } \end{aligned}$
Consider $(1+i \sqrt{3})^{6 m}={ }^{6 m} C_0+{ }^{6 m} C_1(i \sqrt{3})+{ }^{6 m} C_2(i \sqrt{3})^2$
$+{ }^{6 m} C_3(i \sqrt{3})^3+{ }^{6 m} C_4(i \sqrt{3})^4+{ }^{6 m} C_5(i \sqrt{3})^5$
$+\ldots+{ }^{6 m} C_{6 m-1}(i \sqrt{3})^{6 m-1}+{ }^{6 m} C_{6 m}(i \sqrt{3})^{6 m} \ldots(i i)$
Now, $(1+i \sqrt{3})^{6 m}=\left\{(-2)\left(\frac{-1-i \sqrt{3}}{(2)}\right)\right\}^{6 m}=\left(-2 \omega^2\right)^{6 m}$ $=2^{6 m}$, where $\omega$ is cube root of unity.
Then, Eq . (ii) can be written as
$\begin{aligned} & 2^{6 m}=\left\{{ }^{6 m} C_0-{ }^{6 m} C_2 \cdot 3+{ }^{6 m} C_4 \cdot 3^2\right. \\ & \left.-\ldots+(-3)^{3 m} \cdot{ }^{6 m} C_{6 m}\right\}+i \sqrt{3}\left\{{ }^{6 m} C_1-{ }^{6 m} C_3 \cdot 3+\right. \\ & \left.+{ }^{6 m} C_5 \cdot 3^2-\ldots+(-3)^{3 m-1} \cdot{ }^{6 m} C_{6 m-1}\right\} \end{aligned}$
On comparing the imaginary part on both sides, we get
$\sqrt{3}\left({ }^{6 m} C_1-3+{ }^{6 m} C_3+3^2 \cdot{ }^{6 m} C_5-\ldots+(-3)^{3 m-1} \cdot{ }^{6 m} C_{6 m-1}\right)=0$
or ${ }^{6 m} C_1-3+{ }^{6 m} C_3+3^2 \cdot{ }^{6 m} C_5-\ldots+(-3)^{3 m-1} \cdot{ }^{6 m} C_{6 m-1}=0$
$\Rightarrow \sum_{r=1}^{3 m}(-3)^{3 m-1} \cdot{ }^{3 n} C_{2 r-1}=0$
or $\sum_{r=1}^k(-3)^{r-1} \cdot{ }^{3 n} C_{2 r-1}=0$, where $\mathrm{n}=2 \mathrm{~m}$ and $\mathrm{k}=3 \mathrm{~m}$.

Hint

Given, $G=(5 \sqrt{5}+11)^{2 n+1}, 0<G<1$
and $R=(5 \sqrt{5}+11)^{2 n+1}$
$\begin{aligned} & \therefore \quad R-G=2\left\{^{2 n+1} C_1(5 \sqrt{5})^{2 n} \cdot 11+{ }^{2 n+1} C_3\right. \\ &\left.(5 \sqrt{5})^{2 n-2} \cdot(11)^3+\ldots\right\} \end{aligned}$
$\Rightarrow R-G=$ even integer.
or $I+f-G=$ even integer
$f-G=\text { integer } . . \text { (i) }$
where $0 \leq f<1$ and $0<G<1$
$\therefore \quad-\mathrm{i}<f-G<1 \dots(ii)$
From Eqs. (i) and (ii), we get
$\begin{aligned} & f-G=0 \\ & \therefore \quad f=G \\ & \text { Thus, } R \cdot f=R \cdot G \\ &=(5 \sqrt{5}+11)^{2 n+1} \cdot(5 \sqrt{5}-11)^{2 n+1} \\ &= 4^{2 n+1} \end{aligned}$

Hint

$\begin{aligned} & \sum_{r=0}^{20}{ }^{50-r} C_6={ }^{50} C_6+{ }^{49} C_6+{ }^{48} C_6+\ldots . .+{ }^{30} C_6 \\ & ={ }^{50} C_6+{ }^{49} C_6+\ldots . .+{ }^{31} C_6+\left({ }^{30} C_6+{ }^{30} C_7\right)-{ }^{30} C_7 \\ & ={ }^{50} C_6+{ }^{49} C_6+\ldots . .+\left({ }^{31} C_6+{ }^{31} C_7\right)-{ }^{30} C_7 \\ & ={ }^{50} C_6+{ }^{50} C_7-{ }^{30} C_7 \\ & ={ }^{51} C_7-{ }^{30} C_7 \\ & { }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r \end{aligned}$

Hint

(b) Given, ${ }^{20} C_1+2^2 \cdot{ }^{20} C_2+3^2{ }^{20} C_3+\ldots+20^2 \cdot{ }^{20} C_{20}$ $=A\left(2^b\right)$
Taking L.H.S.,
$\begin{aligned} & =\sum_{r=1}^{20} r^2 \cdot{ }^{20} C_r=20 \sum_{r=1}^{20} r \cdot{ }^{19} C_{r-1} \\ & =20\left[\sum_{r=1}^{20}(r-1){ }^{19} C_{r-1}+\sum_{r=1}^{20}{ }^{19} C_{r-1}\right] \\ & =20\left[19 \sum_{r=2}^{20}{ }^{18} C_{r-2}+2^{19}\right]=20\left[19 \cdot 2^{18}+2^{19}\right] \end{aligned}$
$=420 \times 2^{18}$
Now, compare it with R.H.S., $\mathrm{A}=420$ and $\mathrm{b}=18$

Hint

(c) Given expression is $\left(1+a x+b x^2\right)(1-3 x)^{15}$
Co-efficient of $x^2=0$
$\begin{aligned} & \left.\Rightarrow{ }^{15} \mathrm{C}_2(-3)\right)^2+a \cdot{ }^{15} \mathrm{C}_1(-3)+b \cdot{ }^{15} \mathrm{C}_0=0 \\ & \Rightarrow \frac{15 \times 14}{2} \times 9-15 \times 3 a+b=0 \\ & \Rightarrow 945-45 a+b=0 \dots(i) \end{aligned}$
Now, co-efficient of $x^3=0$
$\begin{aligned} & \Rightarrow{ }^{15} \mathrm{C}_3(-3)^3+a \cdot{ }^{15} \mathrm{C}_2(-3)^2+b \cdot{ }^{15} \mathrm{C}_1(-3)=0 \\ & \Rightarrow \frac{15 \times 14 \times 13}{3 \times 2} \times(-3 \times 3 \times 3)+a \times \frac{15 \times 14 \times 9}{2}-b \times 3 \times 15=0 \\ & \Rightarrow 15 \times 3[-3 \times 7 \times 13+a \times 7 \times 3-b]=0 \\ & \Rightarrow 21 a-b=273 \dots(ii) \end{aligned}$
From (i) and (ii), we get,
$a=28, b=315 \Rightarrow(a, b) \equiv(28,315)$

Hint

(b) $2 .{ }^{20} \mathrm{C}_0+5 .{ }^{20} \mathrm{C}_1+8 .{ }^{20} \mathrm{C}_2+\ldots \ldots . .+62 .{ }^{20} \mathrm{C}_{20}$
$\begin{aligned} & =\sum_{r=0}^{20}(3 r+2){ }^{20} C_r=3 \sum_{r=0}^{20} r \cdot{ }^{20} C_r+2 \sum_{r=0}^{20}{ }^{20} C_r \\ & =60 \sum_{r=1}^{20}{ }^{19} C_{n-1}+2 \sum_{r=0}^{20}{ }^{20} C_r \\ & =60 \times 2^{19}+2 \times 2^{20}=2^{21}[15+1]=2^{25} \end{aligned}$

Hint

Let $S={ }^{20} \mathrm{C}_r{ }^{20} \mathrm{C}_0+{ }^{20} \mathrm{C}_{r-1}{ }^{20} \mathrm{C}_1+{ }^{20} \mathrm{C}_{r-2}{ }^{20} \mathrm{C}_2+\ldots+{ }^{20} \mathrm{C}_0{ }^{20} \mathrm{C}_r$ The sum $S$ is the coefficient of $x^r$ in the expansion of $(1+x)^{20}(x+1)^{20}=(1+x)^{40}$ $\therefore \mathrm{S}={ }^{40} \mathrm{C}_{\mathrm{r}}$
$\mathrm{S}$ is maximum when $r=20$

Hint

From question, the summation given is:
$\begin{aligned} & \sum_{\mathrm{r}=0}^{25}\left\{{ }^{50} \mathrm{C}_{\mathrm{r}} \cdot 50-\mathrm{r}_{\mathrm{C}_{25}-\mathrm{r}}\right\}=\mathrm{K}\left({ }^{50} \mathrm{C}_{25}\right) \\ & \because\left[{ }^n C_r=\frac{n !}{r !(n-r) !}\right] \\ & \Rightarrow \sum_{r=0}^{25}\left(\frac{50 !}{r !(50-r) !} \times \frac{(50-r) !}{(25-r) ! 25 !}\right)=K^{50} C_{25} \\ & \Rightarrow \sum_{r=0}^{25}\left(\frac{50 !}{25 ! 25 !} \times \frac{25 !}{r !(25-r) !}\right)=K^{50} C_{25} \end{aligned}$
[On multiplying 25! numerator and denominator]
$\begin{aligned} & \because\left[{ }^{50} C_{25}=\frac{50}{25 ! 25 !}\right] \\ & \Rightarrow{ }^{50} \mathrm{C}_{25} \sum_{r=0}^{25}{ }^{25} \mathrm{C}_r=K^{50} C_{25} \\ & \because\left[{ }^{\mathrm{n}} \mathrm{C}_0+{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} \mathrm{C}_2+\ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}=2^{\mathrm{n}}\right] \\ & \Rightarrow \mathrm{K}=\sum_{\mathrm{r}=0}^{25}{ }^{25} \mathrm{C}_r \\ & \therefore \mathrm{K}=2^{25} \end{aligned}$

Hint

(b) Consider the expression
$\begin{gathered} \left(\frac{1-t^6}{1-t}\right)^3=\left(1-t^6\right)^3(1-t)^{-3} \\ =\left(1-3 t^6+3 t^{12}-t^{18}\right)\left(1+3 t+\frac{3 \cdot 4}{2 !} t^2\right. \\ \left.+\frac{3 \cdot 4 \cdot 5}{3 !} t^3+\frac{3 \cdot 4 \cdot 5 \cdot 6}{4 !} t^4+\ldots \infty\right) \end{gathered}$
Hence, the coefficient of $t^4=1 \cdot \frac{3 \cdot 4 \cdot 5 \cdot 6}{4 !}$
$=\frac{3 \times 4 \times 5 \times 6}{4 \times 3 \times 2 \times 1}=15$

Hint

(a) We have $\left({ }^{21} \mathrm{C}_1+{ }^{21} \mathrm{C}_2 \ldots \ldots . .+{ }^{21} \mathrm{C}_{10}\right)$
$\begin{aligned} & -\left({ }^{10} \mathrm{C}_1+{ }^{10} \mathrm{C}_2 \ldots .{ }^{10} \mathrm{C}_{10}\right) \\ & =\frac{1}{2}\left[\left({ }^{21} \mathrm{C}_1+\ldots .+{ }^{21} \mathrm{C}_{10}\right)+\left({ }^{21} \mathrm{C}_{11}+\ldots .{ }^{21} \mathrm{C}_{20}\right)\right]-\left(2^{10}-1\right) \\ & \left(\because{ }^{10} \mathrm{C}_1+{ }^{10} \mathrm{C}_2+\ldots .+{ }^{10} \mathrm{C}_{10}=2^{10}-1\right) \\ & =\frac{1}{2}\left[2^{21}-2\right]-\left(2^{10}-1\right) \\ & =\left(2^{20}-1\right)-\left(2^{10}-1\right)=2^{20}-2^{10} \\ & \end{aligned}$

Hint

(b) Total number of terms $={ }^{\mathrm{n}+2} \mathrm{C}_2=28$ $(n+2)(n+1)=56 ; n=6$
$\therefore$ Put $x=1$ in expansion $\left(1-\frac{2}{x}+\frac{4}{x^2}\right)^6$,
we get sum of coefficient $=(1-2+4)^6$ $=3^6=729$.

Hint

(c) We know that $(a+b)^n+(a-b)^n$
$\begin{aligned} & =2\left[{ }^n C_0 a^n b^0+{ }^n C_2 a^{n-2} b^2+{ }^n C_4 a^{n-4} b^4 \ldots\right] \\ & (1-2 \sqrt{\mathrm{x}})^{50}+(1+2 \sqrt{\mathrm{x}})^{50} \\ & 2\left[{ }^{50} C_0+{ }^{50} C_2(2 \sqrt{x})^2+{ }^{50} C_4(2 \sqrt{x})^4 \ldots\right] \\ & =2\left[{ }^{50} \mathrm{C}_0+{ }^{50} \mathrm{C}_2 2^2 \mathrm{x}+{ }^{50} \mathrm{C}_4 2^4 \mathrm{x}^2+\ldots\right] \end{aligned}$
Putting $\mathrm{x}=1$, we get,
${ }^{50} C_0+{ }^{50} C_2 2^2+{ }^{50} C_4 2^4 \ldots=\frac{3^{50}+1}{2}$

Hint

(c) Coeff. of $x^{11}$ in exp. of $\left(1+x^2\right)^4\left(1+x^3\right)^7\left(1+x^4\right)^{12}$ $=\left[\right.$ Coeff. of $x^a$ in $\left.\left(1+x^2\right)^4\right] \times\left[\right.$ Coeff. of $x^b$ in $\left.\left(1+x^3\right)^7\right]$
Such that $a+b+c=11$
Here $a=2 m, b=3 n, c=4 p$
$\therefore 2 m+3 n+4 p=11$
Case I : $m=0, n=1, p=2$
Case II : $m=1, n=3, p=0$
Case III : $m=2, n=1, p=1$
Case IV : $m=4, n=1, p=0$
$\therefore$ Required coefficient.
$={ }^4 C_0 \times{ }^7 C_1 \times{ }^{12} C_2+{ }^4 C_1 \times{ }^7 C_3 \times{ }^{12} C_0+{ }^4 C_2 \times{ }^7 C_1 \times{ }^{12} C_1+{ }^4 C_4 \times{ }^7 C_1 \times{ }^{12} C_0$
[Coeff. of $x^c$ in $\left.(1+x)^4\right]$
$=462+140+504+7=1113$

Hint

$\begin{aligned} & (1-x)^{30}={ }^{30} C_0 x^0-{ }^{30} C_1 x^1+{ }^{30} C_2 x^2 \\ & +\ldots \ldots \ldots \ldots+(-1)^{30}{ }^{30} C_{30} x^{30} \ldots \ldots \ldots \ldots(i) \\ & (x+1)^{30}={ }^{30} C_0 x^3 \mathrm{O}-{ }^{30} C_1 x^{29}+{ }^{30} C_2 x^{28} \\ & +\ldots \ldots \ldots \ldots+{ }^{30} C_{10} x^{20}+\ldots \ldots \ldots . .+{ }^{30} C_{30} x^0 \ldots \ldots(ii) \end{aligned}$
Multiplying (i) and (ii) and equating the coefficient of $x^{20}$ on both sides, we get the required sum $=$ coefficient of $x^{20}$ in $\left(1-x^2\right)^{30}={ }^{30} C_{10}$.

Hint

(c) $\begin{aligned} \sum_{i=0}^m{ }^{10} C_i^{20} C_{m-i}={ }^{10} C_0{ }^{20} C_m & +{ }^{10} C_1{ }^{20} C_{m-1} \\ & +{ }^{10} C_2{ }^{20} C_{m-2}+\ldots . .+{ }^{10} C_m{ }^{20} C_0\end{aligned}$ $=$ Coeff. of $x^m$ in the expansion of product $(1+x)^{10}$
$=$ Coeff. of $x^m$ in the expansion of $(1+x)^{30}$ $={ }^{30} C_m$
$\sum_{i=0}^n{ }^{10} C_i{ }^{20} C_{m-1}$ will be maximum, if ${ }^{30} C_m$ will be maximum.
Clearly, ${ }^{30} C_m$ will be maximum when $m=\frac{30}{2}=15$
$\left[\because \operatorname{Max} \cdot\left({ }^n C_r\right)=\left\{\begin{array}{l} { }^n C_{n / 2} \text { if } n \text { is even } \\ { }^n C_{\frac{n+1}{2}} \text { if } n \text { is odd } \end{array}\right]\right.$

Hint

(d) $\left(\begin{array}{l}n \\ r\end{array}\right)+2\left(\begin{array}{c}n \\ r-1\end{array}\right)+\left(\begin{array}{c}n \\ r-2\end{array}\right)$
$\begin{gathered} \left.=\left[\left(\begin{array}{l} n \\ r \end{array}\right)+\left(\begin{array}{c} n \\ r-1 \end{array}\right)\right]+\left[\left(\begin{array}{c} n \\ r-1 \end{array}\right)+\left(\begin{array}{c} n \\ r-2 \end{array}\right)\right\}\right] \\ =\left[\begin{array}{l} \text { Here }\left[\begin{array}{l} n \\ r \end{array}\right],\left[\begin{array}{c} n \\ r-1 \end{array}\right] \text { and }\left[\begin{array}{c} n \\ r-2 \end{array}\right] \\ \text { represent }{ }^n C_r,{ }^n \mathrm{C}_{\mathrm{r}-1} \text { and }{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-2} \end{array}\right] \end{gathered}$
$=\left(\begin{array}{c} n+1 \\ r \end{array}\right)+\left(\begin{array}{c} n+1 \\ r \end{array}\right)=\left(\begin{array}{c} n+2 \\ r \end{array}\right) \quad\left[\because{ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\right]$

Hint

(c) $(1+x)^m(1-x)^n$
$\begin{gathered} =\left[1+m x+\frac{m(m-1)}{2 !} x^2+\ldots\right]\left[1-n x+\frac{n(n-1)}{2 !} x^2-\ldots\right] \\ =1+(m-n) x+\left[\frac{m(m-1)}{2}+\frac{n(n-1)}{2}-m n\right] x^2+\ldots \end{gathered}$
Given, $m-n=3 \dots(i)$
$\text { and } \begin{aligned} & \frac{1}{2} m(m-1)+\frac{1}{2} n(n-1)-m n=-6 \\ & \Rightarrow m^2+n^2-2 m n-(m+n)=-12 \\ & \Rightarrow(m-n)^2-(m+n)=-12 \\ & \Rightarrow m+n=9+12=21 \dots(ii) \end{aligned}$
From (i) and (ii), we get $m=12$

Hint

(c) Given expression:
$\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5$
We know that using binomial theorem,
$(x+a)^n+(x-a)^n=2\left[{ }^n C_0 x^n+{ }^n C_2 x^{n-2} a^2\right.$
$\therefore$ The given expression
$=2\left[{ }^5 C_0 x^5+{ }^5 C_2 x^3\left(x^3-1\right)+{ }^5 C_4 x\left(x^3-1\right)^2\right]$
since maximum power of $x$ involved in the expansionis 7 . Also, only +ve integral powers of $x$ are involved in the expansion, therefore given expression is a polynomial of degree 7.

Hint

(615) General term of the expansion $=\frac{10 !}{\alpha ! \beta ! \gamma !} x^{\beta+2 \gamma}$
For coefficient of $x^4 ; \beta+2 \gamma=4$
Here, three cases arise
Case-1: When $\gamma=0, \beta=4, \alpha=6$
$\Rightarrow \frac{10 !}{\alpha ! \beta ! \gamma !} x^{\beta+2 \gamma}$
Case-2: When $\gamma=1, \beta=2, \alpha=7$
$\Rightarrow \frac{10 !}{7 ! 2 ! 1 !}=360$
Case-3: When $\gamma=2, \beta=0, \alpha=8$
$\Rightarrow \frac{10 !}{8 ! 0 ! 2 !}=45$
Hence, total $=615$

Hint

(30)Let $\left(1-x+x^2 \ldots \ldots x^{2 n}\right)\left(1+x+x^2 \ldots \ldots x^{2 n}\right)$ $=a_0+a_1 x+a_2 x^2+\ldots \ldots$
put $x=1$
$1(2 n+1)=a_0+a_1+a_2+\ldots . a_{2 n} \dots(i)$
put $x=-1$
$1(2 n+1)=a_0+a_1+a_2+\ldots \ldots a_{2 n} \dots(ii)$
Adding (i) and (ii), we get,
$\begin{aligned} 4 n+2 & =2\left(a_0+a_2+\ldots .\right)=2 \times 61 \\ \Rightarrow & 2 n+1=61 \Rightarrow n=30 . \end{aligned}$

Hint

Finding the value of $\sum_{k=0}^n \frac{{ }^n C_k}{k+1}$
Step 1: Consider the given determinant as,
$\left|\begin{array}{cc} \sum_{k=0}^n k & \sum_{k=0}^n{ }^n C_k k^2 \\ \sum_{k=0}^n{ }^n C_k k & \sum_{k=0}^n{ }^n C_k 3^k \end{array}\right|=0$
From the standard formula, we can write the above determinant as,
$\left|\begin{array}{cc} \frac{n(n+1)}{2} & n(n+1) 2^{n-2} \\ n .2^{n-1} & 4^n \end{array}\right|=0$
Step 2: Simplify the above determinant as,
$\begin{aligned} & \Rightarrow \frac{n(n+1)}{2} \cdot 4^n-n^2(n+1) 2^{2 n-3}=0 \\ & \Rightarrow \frac{4^n}{2}-n \frac{4^{n-1}}{2}=0 \\ & \Rightarrow \frac{4^n}{2}=n \frac{4^{n-1}}{2} \\ & \Rightarrow n=\frac{2}{4^{n-1}} \times \frac{4^n}{2} \\ & \Rightarrow n=4^{n-n+1} \\ & \Rightarrow n=4 \end{aligned}$
Step 3: Find the value of $\sum_{k=0}^n \frac{{ }^n C_k}{k+1}$
$\begin{aligned} & \begin{aligned} \sum_{k=0}^n \frac{{ }^n C_k}{k+1} & =\sum_{k=0}^4 \frac{{ }^4 C_k}{k+1} \\ & =\frac{1}{5} \sum_{k=0}^4{ }^5 C_{k+1} \\ & =\frac{1}{5}\left(2^5-1\right) \\ & =\frac{1}{5}(32-1) \\ & =\frac{31}{5} \\ & =6.20 \end{aligned} \\ & \text { Therefore, the value of } \sum_{k=0}^n \frac{{ }^n C_k}{k+1}=6.20 \end{aligned}$

Hint

Sum of coefficients is obtained by putting $x=1$
ie, $(1+1-3)^{2163}=-1$
Thus, sum of the coefficients of the polynomial
$\left(1+x-3 x^2\right)^{2163}$ is $-1$

Hint

$\begin{aligned} & \mathrm{C}_0^2-2 \mathrm{C}_1^2+3 \mathrm{C}_2^2-4 \mathrm{C}_3^2+\ldots+(-1)^{\mathrm{n}}(\mathrm{n}+1) \mathrm{C}_{\mathrm{n}}^2 \\ & =\left[\mathrm{C}_0^2-\mathrm{C}_1^2+\mathrm{C}_2^2-\mathrm{C}_3^2+\ldots+(-1)^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}^2\right]-\left[\mathrm{C}_1^2-2 \mathrm{C}_2^2+3 \mathrm{C}_3^2-\ldots+(-1)^{\mathrm{n}} \mathrm{nC}_{\mathrm{n}}^2\right] \\ & =(-1)^{\frac{n}{2}} \frac{\mathrm{n} !}{\left(\frac{n}{2}\right) !\left(\frac{n}{2}\right) !}-(-1)^{\frac{n}{2}-1} \frac{n}{2} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\frac{n}{2}} \\ & =(-1)^{\frac{n}{2}} \frac{\mathrm{n} !}{\left(\frac{n}{2}\right) !\left(\frac{n}{2}\right) !}\left(1+\frac{n}{2}\right) \\ & \therefore 2\left(\frac{\mathrm{n}}{2}\right) !\left(\frac{\mathrm{n}}{2}\right) !\left[\mathrm{C}_0^2-2 \mathrm{C}_1^2+3 \mathrm{C}_2^2-\ldots+(-1)^{\mathrm{n}}(\mathrm{n}+2) \mathrm{C}_{\mathrm{n}}^2\right] \\ & =(-1)^{\frac{n}{2}}(\mathrm{n}+2) \end{aligned}$

Hint

To show that
$\begin{aligned} & 2^{k, n} C_0 \cdot{ }^n C_k-2^{k-1} \cdot{ }^n C_1 \cdot{ }^{n-1} C_{k-1}+2^{k-2} \cdot{ }^n C_2 \cdot{ }^{n-2} C_{k-2} \\ &-\ldots . .+(-1)^{k n} C_k{ }^{n-k} C_0={ }^n C_k \end{aligned}$
Taking LHS
$\begin{aligned} & 2^k \cdot{ }^n C_0 \cdot{ }^n C_k-2^{k-1} \cdot{ }^n C_1 \cdot{ }^{n-1} C_{k-1}+\ldots+ (-1)^k \cdot{ }^n C_k \cdot{ }^{n-k} C_0\\ &=\sum_{r=0}^k(-1)^r \cdot 2^{k-r} \cdot{ }^n C_r \cdot{ }^{k-r} C_k \cdot{ }^{n-k} C_{k-r} \\ &=\sum_{r=0}^k(-1)^r 2^{k-r} \cdot \frac{n !}{r !(n-r) !} \cdot \frac{(n-r) !}{(k-r) !(n-k) !} \end{aligned}$
$\begin{aligned} & =\sum_{r=0}^k(-1)^r \cdot 2^{k-r} \cdot \frac{n !}{(n-k) ! \cdot k !} \cdot \frac{k !}{r !(k-r) !} \\ & \quad=\sum_{r=0}^k(-1)^r \cdot 2^{k-r}{ }^n C_k \cdot{ }^k C_r=2^k \cdot{ }^n C_k\left\{\sum_{r=0}^k(-1)^r \cdot \frac{1}{2^r} \cdot{ }^k C_r\right\} \\ & \quad=2^k \cdot{ }^n C_k\left(1-\frac{1}{2}\right)^k={ }^n C_k=\text { R.H.S. } \end{aligned}$

Hint

Given that for positive integers $m$ and $n$ such that $n \geq m$, then to prove that
$\begin{aligned} & \begin{array}{c} { }^n C_m+{ }^{n-1} C_m+{ }^{n-2} C_m+\ldots .+{ }^m C_m={ }^{n+1} C_{m+1} \\ \text { L.H.S. }{ }^m C_m+{ }^{m+1} C_m+{ }^{m+2} C_m+\ldots .+{ }^{n-1} C_m+{ }^n C_m \\ \text { [writing L.H.S. in reverse order] } \end{array} \\ & \begin{array}{l} \left({ }^{m+1} C_{m+1}+{ }^{m+1} C_m\right)+{ }^{m+2} C_m+\ldots .+{ }^{n-1} C_m+{ }^n C_m \quad\left[\because \quad{ }^m C_m={ }^{m+1} C_{m+1}\right] \end{array} \\ & =\left({ }^{m+2} C_{m+1}+{ }^{m+2} C_m\right)+{ }^{m+3} C_m+\ldots .+{ }^n C_m \quad\left[\because \quad{ }^n C_{r+1}+{ }^n C_r={ }^{n+1} C_{r+1}\right] \end{aligned}$
Combining in the same way we get
$={ }^n C_{m+1}+{ }^n C_m={ }^{n+1} C_{m+1}=\text { R.H.S. }$
Again we have to prove
$\begin{aligned} & \quad{ }^n C_m+2^{n-1} C_m+3{ }^{n-2} C_m+\ldots+(n-m+1){ }^m C_m \\ & ={ }^{n+2} C_{m+2} \\ & =\left[{ }^n C_m+{ }^{n-1} C_m+{ }^{n-2} C_m+\ldots+{ }^m C_m\right]+\left[{ }^{n-1} C_m\right. \\ & \left.+{ }^{n-2} C_m+\ldots .+{ }^m C_m\right]+\left[{ }^{n-2} C_m+\ldots .+{ }^m C_m\right]+\ldots .+\left[{ }^m C_m\right] \\ & {[n-m+1 \text { bracketed terms }]} \\ & ={ }^{n+1} C_{m+1}+{ }^n C_{m+1}{ }^{n-1} C_{m+1} \ldots+{ }^{m+1} C_{m+1} \\ & ={ }^{n+2} C_{m+2} \\ & {[\text { Replacing } n \text { by } n+1 \text { and } m \text { by } m+1 \text { in the previous result.] }} \\ & =\text { R.H.S. } \end{aligned}$

Hint

$\text { Given : }\left(1+x+x^2\right)^n=a_0+a_1 x+\ldots .+\mathrm{a}_{2 n} x^{2 n} \dots(i)$
where $n$ is a +ve integer.
On replacing $\mathrm{x}$ by $-\frac{1}{x}$ in equation(i), we get
$\left(1-\frac{1}{x}+\frac{1}{x^2}\right)^n=a_0-\frac{a_1}{x}+\frac{a_2}{x^2}-\frac{a_3}{x^3}+\ldots .+\frac{a_{2 n}}{x^{2 n}} \dots(ii)$
Multiplying equation (i) and (ii) :
$\begin{gathered} \frac{\left(1+x+x^2\right)^n\left(x^2-x+1\right)^n}{x^{2 n}} \\ =\left(a_0+a_1 x+\ldots .+a_{2 n} x^{2 n}\right)\left(a_0-\frac{a_1}{x}+\frac{a_2}{x^2}+\ldots+\frac{a_{2 n}}{x^{2 n}}\right) \end{gathered}$
Equating the constant terms on both sides we get
$a_0^2-a_1^2+a_2^2-a_3^2+\ldots .+a_{2 n}^2=$ constant term in the expansion of
$\frac{\left[\left(1+x+x^2\right)\left(1-x+x^2\right)\right]^n}{x^{2 n}}$
$=$ Coeff. of $x^{2 n}$ in the expansion of $\left(1+x^2+x^4\right)^n$
But replacing $x$ by $x^2$ in equation (i), we have
$\begin{aligned} & \left(1+x^2+x^4\right)^n=a_0+a_1 x^2+\ldots+a_{2 n}\left(x^2\right)^{2 n} \\ & \quad \therefore \quad \text { Coeff. of } x^{2 n}=a_n \\ & \therefore \quad a_0^2-a_1^2+a_2^2-a_3^2+\ldots .+a_{2 n}^2=a_n \end{aligned}$

Hint

Given : $\sum_{r=0}^{2 n} a_r(x-2)^r=\sum_{r=0}^{2 n} b_r(x-3)^r \dots(i)$
and $a_k=1, \forall k \geq n$
To prove $: b_n={ }^{2 n+1} C_{n+1}$
In the given equation (i) let us put $x-3=y$
$\begin{aligned} & \Rightarrow x-2=y+1 \\ & \therefore \sum_{r=0}^{2 n} a_r(1+y)^r=\sum_{r=0}^{2 n} b_r(y)^r[\text { From (i) }] \\ & \Rightarrow a_0+a_1(1+y)+\ldots .+a_{n-1}(1+y)^{n-1}+(1+y)^n \\ & +(1+y)^{n+1}+\ldots .+(1+y)^{2 n} \end{aligned}$
$=\sum_{r=0}^{2 n} b_r y^r$
[Using $a_k=1, \forall k \geq n$ ]
Equating the coefficients of $y^n$ on both sides we get
$\begin{aligned} & \Rightarrow{ }^n C_n+{ }^{n+1} C_n+{ }^{n+2} C_n+\ldots .+{ }^{2 n} C_n=b_n \\ & \Rightarrow\left({ }^{n+1} C_{n+1}+{ }^{n+1} C_n\right)+{ }^{n+2} C_n+\ldots .+{ }^{2 n} C_n=b_n \end{aligned}$
$\left[{ }^n C_n={ }^{n+1} C_{n+1}=1\right]$
$\Rightarrow b_n={ }^{n+2} C_{n+1}+{ }^{n+2} C_n+\ldots .+{ }^{2 n} C_n$
$\left[{ }^m C_r+{ }^m C_{r-1}={ }^{m+1} C_r\right]$
Combining the terms in similar way, we get
$\Rightarrow b_n={ }^{2 n} C_{n+1}+{ }^{2 n} C_n \Rightarrow \quad b_n={ }^{2 n+1} C_{n+1}$

Hint

We know
$(1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots .+(-1)^n C_n x^n$
On multiplying both sides by $x$,
$x(1-x)^n=C_0 x-C_1 x^2+C_2 x^3-C_3 x^4+\ldots .+(-1)^n C_n x^{n+1}$
On differentiating both sides w.r. to $x$,
$\begin{aligned} & (1-x)^n-n x(1-x)^{n-1} \\ & =C_0-2 C_1 x+3 C_2 x^2-4 C_3 x^3+\ldots .+(-1)^n(n+1) C_n x^n \end{aligned}$
Again on multiplying both sides by $x$,
$\begin{gathered} x(1-x)^n-n x^2(1-x)^{n-1} \\ =C_0 x-2 C_1 x^2+3 C_2 x^3-4 C_3 x^4+\ldots .+(-1)^n(n+1) C_n x^{n+1} \end{gathered}$
On differentiating both sides with respect to $x$,
$\begin{aligned} & (1-x)^n-n x(1-x)^{n-1}-2 n x(1-x)^{n-1}+n x^2(n-1)(1-x)^{n-2} \\ & =C_0-2^2 C_1 x+3^2 C_2 x^2-4^2 C_3 x^3+\ldots .+(-1)^n(n+1)^2 C_n x^n \end{aligned}$
Putting $x=1$, in above, we get
$0=C_0-2^2 C_1+3^2 C_2-4^2 C_3+\ldots .+(-1)^n(n+1)^2 C_n$

Hint

$\begin{aligned} & { }^{n+1} C_1+{ }^{n+1} C_2 s_1+{ }^{n+1} C_3 s_2+\ldots .+{ }^{n+1} C_{n+1} s_n \\ & =\sum_{r=1}^{n+1}{ }^{n+1} C_r s_{r-1}, \end{aligned}$
where
$\begin{aligned} & s_n=1+q+q^2+\ldots+q^n=\frac{1-q^{n+1}}{1-q} \\ & \therefore \sum_{r=1}^{n+1}{ }^{n+1} C_r\left(\frac{1-q^r}{1-q}\right)=\frac{1}{1-q}\left(\sum_{r=1}^{n+1}{ }^{n+1} C_r-\sum_{r=1}^{n+1}{ }^{n+1} C_r q^r\right) \\ & =\frac{1}{1-q}\left[(1+1)^{n+1}-(1+q)^{n+1}\right] \\ & =\frac{1}{1-q}\left[2^{n+1}-(1+q)^{n+1}\right] \quad \ldots \text { (i) } \end{aligned}$
Also, $S_n=1+\left(\frac{q+1}{2}\right)+\left(\frac{q+1}{2}\right)^2+\ldots+\left(\frac{q+1}{2}\right)^n$
$=\frac{1-\left(\frac{q+1}{2}\right)^{n+1}}{1-\left(\frac{q+1}{2}\right)}=\frac{2^{n+1}-(q+1)^{n+1}}{2^n(1-q)} \dots(ii)$
From equations (i) and (ii),
${ }^{n+1} C_1+{ }^{n+1} C_2 s_1+{ }^{n+1} C_3 s_2+\ldots+{ }^{n+1} C_{n+1} s_n=2^n S_n$

Hint

$\begin{aligned} & S=\sum \sum C_i C_j \\ & 0 \leq i<j \leq n \\ & \Rightarrow S=C_0\left(C_1+C_2+C_3+\ldots .+C_n\right)+C_1\left(C_2+C_3+\ldots .+C_n\right) \\ & +C_2\left(C_3+C_4+C_5+\ldots C_n\right)+\ldots C_{n-1}\left(C_n\right) \\ & \Rightarrow S=C_0\left(2^n-C_0\right)+C_1\left(2^n-C_0-C_1\right)+ \\ & C_2\left(2^n-C_0-C_1-C_2\right) \\ & +\ldots .+C_{n-1}\left(2^n-C_0-C_1 \ldots . C_{n-1}\right) \\ & +C_n\left(2^n-C_0-C_1 \ldots C_n\right) \\ & \Rightarrow S=2^n\left(C_0+C_1+C_2+\ldots .+C_{n-1}+C_n\right) \\ & -\left(C_0^2+C_1^2+C_2^2+\ldots .+C_n^2\right)-S \\ & \Rightarrow 2 S=2^n \cdot 2^n-\frac{2 n !}{(n !)^2}=2^{2 n}-\frac{2 n !}{(n !)^2} \\ & \end{aligned}$

Hint

Given : $C_1+2 C_2 x+3 C_3 x^2+\ldots . .+2 n C_{2 n} \pi^{2 n-1}=2 n(1+x)^{2 n-1} \ldots .(\mathrm{i})$
where $C_r=\frac{2 n !}{r !(2 n-r) !}$
Integrating both sides with respect to $x$, under the limits 0 to $x$, we get
$\begin{aligned} & {\left[C_1 x+C_2 x^2+C_3 x^3+\ldots .+C_{2 \pi} x^{2 n}\right]{ }_0^x=\left[(1+x)^{2 n}\right]_0^x} \\ & \quad \Rightarrow C_1 x+C_2 x^2+C_3 x^3+\ldots .+C_{2 n} x^{2 n}=(1+x)^{2 n}-1 \\ & \quad \Rightarrow C_0+C_1 x+C_2 x^2+C_3 x^3+\ldots .+C_{2 \pi} x^{2 n}=(1+x)^{2 n} \dots(ii) \end{aligned}$
Changing $x$ by $-\frac{1}{x}$, we get
$\begin{aligned} & \Rightarrow C_0-\frac{C_1}{x}+\frac{C_2}{x^2}-\frac{C_3}{x^3}+\ldots .+(-1)^{2 n} \frac{C_{2 n}}{x^{2 n}}=\left(1-\frac{1}{x}\right)^{2 n} \\ & \Rightarrow C_0 x^{2 n}-C_1 x^{2 n-1}+C_2 x^{2 n-2}-C_3 x^{2 n-3} \\ & +\ldots . .+C_{2 n}=(x-1)^{2 n} \dots(iii) \end{aligned}$
Multiplying eqn. (i) and (iii) and equating the coefficients of $\mathrm{x}^{2 \mathrm{n}-1}$ on both sides, we get
$\begin{aligned} & \quad C_1^2+2 C_2^2-3 C_3^2+\ldots+2 n C_{2 n}^2 \\ &=\text { coeff.of } x^{2 n-1} \text { in } 2 n(x-1)\left(x^2-1\right)^{2 n-1} \\ &=2 n \text { [coeff. of } x^{2 n-2} \text { in }\left(x^2-1\right)^{2 n-1} \left.-\text { coeff. of } x^{2 n-1} \text { in }\left(x^2-1\right)^{2 n-1}\right]\\ &=2 n\left[\left[^{2 n-1} C_{n-1}(-1)^{n-1}-0\right]\right. \\ &=(-1)^{n-1} \cdot 2 n^{2 n-1} C_{n-1} \\ & \quad \Rightarrow\left.-\text { coeff. of } x^{2 n-1} \text { in }\left(x^2-1\right)^{2 n-1}\right] \\ &=(-1)^n \cdot 2 n^{2 n-1} C_{n-1}=(-1)^n n \cdot\left(\frac{2 n}{n} \cdot{ }^{2 n-1} C_{n-1}\right) \\ &=(-1)^n n \cdot{ }^{2 n} C_n=(-1)^n n \cdot C_n . \quad\left(\because 3 C_3^2+\ldots .+2 n C_n C_n\right) \end{aligned}$

Hint

(a)

$\begin{aligned} & 428=21 \times 20+8 \\ \Rightarrow \quad & (428)^{2024} =(20 \times 21+8)^{2024} = 8^{2024}(\bmod 21) \\ & 8^2=21 \times 3+1 \\ & 8^{2024}=(21 \times 3+1)^{1012} \\ \Rightarrow \quad & 8^{2024} =(21 \times 3+1)^{1012}(\bmod 21) \\ & = 1^{2012}(\bmod 21) \\ & 428^{2024} = 1(\bmod 21) \end{aligned}$

Hint

$\begin{aligned} & T_2={ }^n C_1 y^1 \cdot x^{n-1}=135 \dots(i)\\ & T_3={ }^n C_2 y^2 \cdot x^{n-2}=30 \dots(ii) \\ & T_4={ }^n C_3 y^3 x^{n-3}=\frac{10}{3} \dots(iii) \end{aligned}$
By $\frac{\text { (i) }}{\text { (ii) }}$
$\frac{{ }^n C_1}{{ }^n C_2} \frac{x}{y}=\frac{9}{2} \dots(iv)$
$\begin{aligned} & \text { By } \frac{\text { (ii) }}{\text { (iii) }} \\ & \frac{{ }^n C_2}{{ }^n C_3} \frac{x}{y}=9 \dots(v) \\ & \end{aligned}$
$\begin{aligned} & \text { By } \frac{( iv )}{( v )} \\ & \frac{{ }^n C_1{ }^n C_3}{{ }^n C_2{ }^n C_2}=\frac{1}{2} \\ & \frac{2 n ^2( n -1)( n -2)}{6}=\frac{ n ( n -1)}{2} \frac{ n ( n -1)}{2} \\ & 4 n-8=3 n-3 \\ & \Rightarrow n =5 \\ & \text { put in (v) } \\ & \frac{x}{y}=9 \\ & x=9 y \\ & \text { put in (i) } \\ & { }^5 C_1 x^4\left(\frac{x}{9}\right)=135 \\ & x^5=27 \times 9 \\ & \Rightarrow x=3, \quad y=\frac{1}{3} \\ & 6\left(n^3+x^2+y\right) \\ & =6\left(125+9+\frac{1}{3}\right) \\ & =806 \end{aligned}$

Hint

General term of $\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$
$T_{r+1}={ }^9 C_r\left(\frac{3}{2} x^2\right)^{9-r}\left(-\frac{1}{3 x}\right)^r={ }^9 C_r(-1)^r 3^{9-2 r} 2^{r-9} x^{18-35}$
Constant term in expansion of $\left(1+2 x-3 x^3\right)$
$\begin{aligned} & \left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 \\ & =T_7-3 T_8={ }^9 C_6 3^{-3} \cdot 2^{-3}+3^9 C_7 \cdot 3^{-5} \cdot 2^{-2} \\ & =\frac{3 \times 4 \times 7}{3^3 \cdot 2^3}+\frac{3 \times 9 \times 4}{3^5 \times 2^2}= \\ & p=\frac{42+12}{108}=\frac{54}{108} \\ & 108 p=54 \end{aligned}$

Hint

$\begin{aligned} & a=1+\frac{{ }^2 C_2}{3!}+\frac{{ }^3 C_2}{4!}+\frac{{ }^4 C_2}{5!}+\ldots \\ & b=1+\frac{{ }^1 C_0+{ }^1 C_1}{1!}+\frac{{ }^2 C_0+{ }^2 C_1+{ }^2 C_2}{2!}+\ldots \\ & b=1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2}{3!}+\ldots=e^2 \\ & \text { Using } e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x}{3!}+\ldots \\ & a=1+\sum_{r=2}^{\infty} \frac{{ }^r C_2}{(r+1)!}=1+\sum_{r=2} \frac{r(r-1)}{2(r+1)!} \\ & =1+\frac{1}{2} \sum_{r=2}^{\infty} \frac{(r+1) r-2 r}{(r+1)!} \\ & =1+\frac{1}{2} \sum_{r=2}^{\infty} \frac{1}{(r-1)!}-\frac{1}{2} \sum_{r=2} \frac{2 r}{(r+1)!} \\ & =1+\frac{1}{2}\left(\frac{1}{1!}+\frac{1}{2!}+\ldots\right)-\sum_{r=2}^{\infty} \frac{(r+1)-1}{(r+1)!} \\ & =1+\frac{1}{2}(e-1)-\sum_{r=2}^{\infty} \frac{1}{r!}+\sum_{r=2} \frac{1}{(r+1)!} \\ & =1+\frac{1}{2}(e-1)-\left(e-\frac{1}{1!}-\frac{1}{0!}\right)+\left(e-\frac{1}{1!}-\frac{1}{0!}-\frac{1}{2!}\right) \\ & =1+\frac{e}{2}-\frac{1}{2}-e+2+e-2-\frac{1}{2}=\frac{e}{2} \\ & \Rightarrow \frac{2 b}{a^2}=\frac{2 {e^2}}{\frac{e^2}{4}}= 8 \\ & \end{aligned}$

Hint

(b) We are given the product of three binomial expansions:
$\left(1+\frac{1}{x}\right)^6\left(1+x^2\right)^7\left(1-x^3\right)^8$
We need to find the coefficient of $x^{30}$ in the expansion of this product.
Step 1: Expanding each binomial term
1. Expand $\left(1+\frac{1}{x}\right)^6$ : The general term for $\left(1+\frac{1}{x}\right)^6$ is:
2. Expand $\left(1+x^2\right)^7$ : The general term for $\left(1+x^2\right)^7$ is:
3. Expand $\left(1-x^3\right)^8$ : The general term for $\left(1-x^3\right)^8$ is:
Step 2: Finding the coefficient of $x^{30}$
Now, we need to find the values of $r, s$, and $t$ such that the exponents of $x$ from all three expansions sum to 30 :
$-r+2 s+3 t=30$
We need to solve for $r, s$, and $t$ such that this equation holds.
Case 1: $r=6$
For $r=6$, we have:
$2 s+3 t=36$
Solving this equation for integer values of $s$ and $t$, we get: $s=12, t=8$. The corresponding terms are:
$\binom{6}{6} \times\binom{ 7}{12} \times\binom{ 8}{8}=678$
Thus, the coefficient $\alpha$ is 678 .

Hint

$\begin{aligned} & (x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3} \\ & (x+2)^2+\ldots \ldots+(x+2)^{n-1} \\ & \sum \alpha_r=4^{n-1}+4^{n-2} \times 3+4^{n-3} \times 3^2 \ldots \ldots+3^{n-1} \\ & =4^{n-1}\left[1+\frac{3}{4}+\left(\frac{3}{4}\right)^2 \ldots+\left(\frac{3}{4}\right)^{n-1}\right] \\ & =4^{n-1} \times \frac{1-\left(\frac{3}{4}\right)^n}{1-\frac{3}{4}} \\ & =4^n-3^n=\beta^n-\gamma^n \\ & \beta=4, \gamma=3 \\ & \beta^2+\gamma^2=16+9=25 \end{aligned}$

Hint

$\begin{aligned} & (1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5 \\ & =(1+x)\left(1-x^2\right)\left(\left(1+\frac{1}{x}\right)^3\right)^5 \\ & =\frac{(1+x)^2(1-x)(1+x)^{15}}{x^{15}} \\ & =\frac{(1+x)^{17}-x(1+x)^{17}}{x^{15}} \end{aligned}$
$=\operatorname{coeff}\left( x ^3\right)$ in the expansion $\approx \operatorname{coeff}\left( x ^{18}\right)$ in
$\begin{aligned} & (1+x)^{17}-x(1+x)^{17} \\ & =0-1 \\ & =-1 \end{aligned}$
coeff $\left( x ^{-13}\right)$ in the expansion $\approx \operatorname{coeff}\left( x ^2\right)$ in
$\begin{aligned} & (1+x)^{17}-x(1+x)^{17} \\ & =\binom{17}{2}-\binom{17}{1} \\ & =17 \times 8-17 \\ & =17 \times 7 \\ & =119 \end{aligned}$
Hence Answer $=119-1=118$

Hint

$\begin{aligned} \alpha= & \sum_{k=0}^n \frac{{ }^n C_k \cdot{ }^n C_k}{k+1} \cdot \frac{n+1}{n+1} \\ & =\frac{1}{n+1} \sum_{k=0}^n{ }^{n+1} C_{k+1} \cdot{ }^n C_{n-k} \\ \alpha & =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+1} \\ \beta & =\sum_{k=0}^{n-1} C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \frac{n+1}{n+1} \\ & \frac{1}{n+1} \sum_{k=0}^{n-1}{ }^n C_{n-k} \cdot{ }^{n+1} C_{k+2} \\ & =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+2} \\ \frac{\beta}{\alpha} & =\frac{2 n+1}{2 n+1} C_{n+2} \\ \frac{\beta}{\alpha} & =\frac{2 n+1-(n+2)+1}{n+2}=\frac{5}{6} \\ n & =10 \end{aligned}$

Hint

General term in expansion of $\left((7)^{1 / 2}+(11)^{1 / 6}\right)^{824}$ is
$t _{ r +1}={ }^{824} C _{ r }(7)^{\frac{824- r }{2}}(11)^{ r / 6}$
For integral term, $r$ must be multiple of 6 .
Hence $r=0,6,12, \ldots \ldots .822$
Total = 138.

Hint

Let $32^{32}= t$
$\begin{aligned} & 64^{32^{32}}=64^t=8^{2 t}=(9-1)^{2 t} \\ & =9 k +1 \end{aligned}$
Hence remainder $=1$

Hint

$\begin{aligned} & \sum_{ r =1}^9 \frac{{ }^{11} C _{ r }}{ r +1} \\ & =\frac{1}{12} \sum_{ r =1}^9{ }^{12} C _{ r +1} \\ & =\frac{1}{12}\left[2^{12}-26\right]=\frac{2035}{6} \\ & \therefore m + n =2041 \end{aligned}$

Hint

$\begin{aligned} & (1-x)(1-x)^{2007}\left(1+x+x^2\right)^{2007} \\ & (1-x)\left(1-x^3\right)^{2007} \\ & (1-x)\left({ }^{2007} C_0-{ }^{2007} C_1\left(x^3\right)+\ldots \ldots\right) \end{aligned}$
General term
$\begin{aligned} & (1-x)\left((-1)^{r 2007} C_r x^{3 r}\right) \\ & (-1)^{r 2007} C_r x^{3 r}-(-1)^{r 2007} C_r x^{3 r+1} \\ & 3 r=2012 \\ & r \neq \frac{2012}{3} \\ & 3 r+1=2012 \\ & 3 r=2011 \\ & r \neq \frac{2011}{3} \end{aligned}$
Hence there is no term containing $x ^{2012}$.
So coefficient of $x^{2012}=0$

Hint

$\begin{aligned} & 7^{103}=7 \times 7^{102} \\ & =7 \times(49)^{51} \\ & =7 \times(51-2)^{51} \\ & \text { Remainder }=7 \times(-2)^{51} \\ & =-7\left(2^3 \cdot(16)^{12}\right) \\ & =-56(17-1)^{12} \\ & \text { Remainder }=-56 \times(-1)^{12}=-56+68=12 \end{aligned}$

Hint

Given expression $\left(\sqrt{x}-\frac{6}{x^{3 / 2}}\right)^n$. Here, $a=\sqrt{x}$ and $b=-\frac{6}{x^{3 / 2}}$.
The $r$-th term of the binomial expansion of $(a+b)^n$ is given by
$T_r={ }^n C_r a^{n-r} b^r \text {. }$
Substitute $a$ and $b$ in this formula, we get:
$T_r={ }^n C_r(\sqrt{x})^{n-r}\left(-\frac{6}{x^{3 / 2}}\right)^r={ }^n C_r(-6)^r x^{\frac{n-4 r}{2}} .$
The constant term in the binomial expansion is obtained when the power of $x$ in the terms equals zero.
This happens when $\frac{n-4 r}{2}=0$, which gives $n=4 r$.
$\begin{aligned} & { }^n C_{\frac{n}{4}}(-6)^{\frac{n}{4}}=\alpha \\ & (-5)^n-{ }^n C_{\frac{n}{4}}(-6)^{n / 4}=649 \end{aligned}$
By observation $(625+24=649)$, we get $n=4$
$\therefore \alpha=-24$
Now, for coefficient of $x^{-4}$
$\begin{aligned} & \frac{n-4 r}{2}=-4 \\ & n=4 r-8 \Rightarrow r=3 \\ & \lambda(-24)=(-6)^3 \cdot{ }^4 C_3 \\ & \Rightarrow \lambda=36 \end{aligned}$

Hint

We have, binomial coefficient, $(2+x)^9$
$T_{r+1}={ }^n C_r 2^{n-r} \times x^r$
Coefficient of $x\left(T_1\right)={ }^9 C_1 \times 2^8$
Coefficient of $x^2\left(T_2\right)={ }^9 C_2 \times 2^7$
Coefficient of $x^3\left(T_3\right)={ }^9 C_3 \times 2^6$
. .
. .
. .
Coefficient of $x^7\left(T_7\right)={ }^9 C_7 \times 2^2$
Mean $=\frac{{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots+{ }^9 C_7 \times 2^2}{7}$
$\begin{aligned} & { }^9 C_0 \times 2^9+{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots+{ }^9 C_7 \times 2^2 \\ & =\frac{+{ }^9 C_8 \times 2^1+{ }^9 C_9 \times 2^0-{ }^9 C_0 \times 2^9-{ }^9 C_8 \times 2^1-{ }^9 C_9 \times 2^0}{7} \\ & =\frac{(1+2)^9-2^9-18-1}{7}=\frac{19152}{7}=2736 \end{aligned}$

Hint

General term of the expansion $\left(3^{\frac{1}{2}}+5^{\frac{1}{4}}\right)^{680}$
$={ }^{680} C_r\left(3^{1 / 2}\right)^{680-r}\left(5^{1 / 4}\right)^r={ }^{680} C_r \times 3^{\frac{680-r}{2}} \times 5^{\frac{r}{4}}$
The term will be integral if $r$ is a multiple of 4 .
$\therefore r=0,4,8,12, \ldots, 680$ (which is an AP)
$\begin{aligned} & 680=0+(n-1) 4 \\ & n=\frac{680}{4}+1=171 \end{aligned}$

Hint

Given expression is $\left(1-x+2 x^3\right)^{10}$
So, general term is $\frac{10!}{r_{1}!r_{2}!r_{3}!}(1)^{r_1}(-1)^{r_2} \cdot(2)^{r_3} \cdot(x)^{r_2+r_3}$
Where, $r_1+r_2+r_3=10$ and $r_2+3 r_3=7$
$\begin{aligned} &\text { Now, for possibility, }\\ &\begin{array}{ccc} r_1 & r_2 & r_3 \\ 3 & 7 & 0 \\ 7 & 1 & 2 \\ 5 & 4 & 1 \end{array} \end{aligned}$
Thus, required co-efficient
$\begin{aligned} & =\frac{10!}{3!7!}(-1)^7+\frac{10!}{5!4!}(-1)^4(2)+\frac{10!}{7!2!}(-1)^1(2)^2 \\ & =-120+2520-1440 \\ & =2520-1560=960 \end{aligned}$

Hint

Let $T _{r+1}$ be the constant term.
$T _{r+1}={ }^7 C _r\left(3 x^2\right)^{7-r}\left(\frac{-1}{2 x^5}\right)^r$
For constant term, power of $x$ should be zero.
i.e., $14-2 r-5 r=0$
$\Rightarrow 14=7 r \Rightarrow r=2$
Now, constant term $=\alpha$
$\begin{aligned} & \Rightarrow{ }^7 C_2(3)^5\left(\frac{-1}{2}\right)^2=\alpha \\ & \Rightarrow 21 \times 243 \times \frac{1}{4}=\alpha \\ & \Rightarrow[\alpha]=[1275.75]=1275 \end{aligned}$

Hint

We have,
$\begin{aligned} & {\left[\frac{66}{3}\right]=22} \\ & {\left[\frac{66}{3^2}\right]=7} \\ & {\left[\frac{66}{3^3}\right]=2} \end{aligned}$
Highest powers of 3 is greater than 66 . So, their g.i.f. is always 0 .
$\therefore$ Required natural number $=22+7+2=31$

Hint

$\begin{aligned} T_{r+1} & ={ }^{15} C_r\left(x^4\right)^{15-r}\left(-\frac{1}{x^3}\right)^r={ }^{15} C_r(-1)^r x^{60-4 r-3 r} \\ & ={ }^{15} C_r(-1)^r x^{60-7 r} \end{aligned}$
$\begin{array}{r} \therefore 60-7 r=18 \\ \Rightarrow 7 r=42 \\ \Rightarrow r=6 \end{array}$
$\therefore$ The coefficient of $x^{18}$
$={ }^{15} C_6(-1)^6=\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{6 \times 5 \times 4 \times 3 \times 2 \times 1}=5005$

Hint

${ }^m C_1,{ }^m C_2,{ }^m C_3$ are first, third and fifth term of $A P$
$\begin{aligned} \therefore \quad & a={ }^m C_1 \\ & a+2 d={ }^m C_2 \\ & a+4 d={ }^m C_3 \end{aligned}$
$\begin{gathered} \therefore \quad 2{ }^m C_2-{ }^m C_3=m \\ \Rightarrow m=7 \text { or } m=2 \end{gathered}$
$\because m=2$ is not possible
$\therefore m=7$
$T _6={ }^{ m } C _5\left(10-3^{ x }\right)^{\frac{ m -5}{2}} \cdot\left(3^{ x -2}\right)=21$
Putting value of $m=7$, we get
$\begin{aligned} & T_{5+1}={ }^7 C_5\left(10-3^x\right)^{\frac{7-5}{2}} 3^{x-2}=21 \\ & \Rightarrow \frac{10.3^x-\left(3^x\right)^2}{3^2}=1 \\ & \Rightarrow\left(3^x\right)^2-10 \cdot 3^x+9=0 \\ & \Rightarrow 3^x=9,1 \\ & \Rightarrow x=0,2 \end{aligned}$
Sum of squares of values of $x=0^2+2^2=4$

Hint

Given expansion $\left(x^{\frac{2}{3}}+\frac{\alpha}{x^3}\right)^{22}$
$T_{r+1}={ }^{22} C_r\left(x^{\frac{2}{3}}\right)^{22-r}\left(\frac{\alpha}{x^3}\right)^r$
For constant term
$\begin{aligned} & \frac{44-2 r}{3}-3 r=0 \\ & \Rightarrow r=4 \end{aligned}$
Now ${ }^{22} C _4 \alpha^4=7315$
$\frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} \alpha^4=7315$
$\begin{aligned} & \therefore \alpha^4=1 \\ & \therefore|\alpha|=1 \end{aligned}$

Hint

$\begin{aligned} & 19^{200}+23^{200} \\ & =(21-2)^{200}+(21+2)^{200}=49 \lambda+2^{201} \end{aligned}$
Now, $2^{201}=8^{67}=(7+1)^{67}=49 \lambda+7 \times 67+1$
$\begin{aligned} & =49 \lambda+470 \\ & =49(\lambda+9)+29 \\ & \therefore \text { Remainder }=29 \end{aligned}$

Hint

Coeff of $x^{-6}$ in $\left(\frac{4 x}{5}+\frac{5}{2 x^2}\right)^9$
$\begin{aligned} T_{r+1} & ={ }^9 C_r\left(\frac{4 x}{5}\right)^{9-r}\left(\frac{5}{2 x^2}\right)^r \\ 9-3 r & =-6 \\ r & =5 \end{aligned}$
Coeff of $x^{-6}={ }^9 C_5\left(\frac{4}{5}\right)^4\left(\frac{5}{2}\right)^5=5040$

Hint

Given binomial expansion of $\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^l}\right)^9$
$\begin{aligned} T_{r+1} & ={ }^9 C_r\left(\frac{x^{\frac{5}{2}}}{2}\right)^{9-r}\left(\frac{-4}{x^l}\right)^r \\ & ={ }^9 C_r x^{\frac{45-5 r}{2}-l r} \cdot 2^{r-9} \cdot 4^r \cdot(-1)^r \end{aligned}$
For constant term, power of $x$ is zero.
So, $\frac{45-5 r}{2}=l r \Rightarrow 2 l r+5 r=45$
Now constant term $=-84$
and ${ }^9 C_r \cdot 2^{3 r-9}(-1)^r=-84$
So, $r=3$ and $l=5$
Now for $x^{-15}, \frac{45-5 r}{2}-5 r=-15$
$\begin{aligned} & \Rightarrow 45-15 r=-30 \\ & \Rightarrow r=5 \end{aligned}$
$\therefore$ Coefficient $=-{ }^9 C_5 2^6=-63.2^7$
$\therefore \alpha=7, \beta=-63$
and $|\alpha l-\beta|=|7 \times 5+63|=98$

Hint

$\begin{aligned} & 5^{99}=5^4 \cdot 5^{95} \\ & =625\left[5^5\right]^{19} \\ & =625[3125]^{19} \\ & =625[3124+1]^{19} \\ & =625[11 k \times 19+1] \\ & =625 \times 11 k \times 19+625 \\ & =11 k _1+616+9 \\ & =11\left( k _2\right)+9 \\ & \text { Remainder }=9 \end{aligned}$

Hint

$\begin{aligned} & T _{ r +1}={ }^{30} C _{ r }\left( x ^{2 / 3}\right)^{30- r }\left(\frac{2}{ x ^3}\right)^{ r } \\ & ={ }^{30} C _{ r } \cdot 2^{ r } \cdot x ^{\frac{60-11 r }{3}} \\ & \frac{60-11 r }{3}<0 \\ & \Rightarrow 11 r >60 \\ & \Rightarrow r >\frac{60}{11} \\ & \Rightarrow r =6 \\ & T _7={ }^{30} C _6 \cdot 2^6 x ^{-2} \end{aligned}$
We have also observed $\beta={ }^{30} C _6(2)^6$ is a natural number.
$\therefore \alpha=2$

Hint

$\begin{aligned} & \text { Given } x^{\frac{1}{50}}=12 \Rightarrow x=12^{50} \\ & y^{\frac{1}{50}}=18 \Rightarrow y=18^{50} \\ & 12 \equiv 13(\operatorname{Mod} 25) \\ & 12^2 \equiv 19(\operatorname{Mod} 25) \\ & 12^3 \equiv-3(\operatorname{Mod} 25) \\ & 12^9 \equiv-2(\operatorname{Mod} 25) \\ & 12^{10} \equiv-1(\operatorname{Mod} 25) \\ & 12^{50} \equiv-1(\operatorname{Mod} 25) \dots(i) \end{aligned}$
Now
$18 \equiv 7(\operatorname{Mod} 25)$
$18^2 \equiv-1(\operatorname{Mod} 25)$
$18^{-50} \equiv-1(\operatorname{Mod} 25) \dots(ii)$
$\therefore 12^{50}+18^{50} \equiv-2(\operatorname{Mod} 25)$
$\equiv 23(\operatorname{Mod} 25)$
$\therefore$ Answer $=23$

Hint

$\begin{aligned} & t _{ r +1}={ }^{ n } C _{ r }(2 x )^{ r } \\ & \Rightarrow \frac{{ }^{ n } C _{ r -1}(2)^{ r -1}}{{ }^{ n } C _{ r }(2)^{ r }}=\frac{2}{5} \\ & \Rightarrow \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!(2)}{r!(n-r)!}}=\frac{2}{5} \\ & \Rightarrow \frac{ r }{ n – r +1}=\frac{4}{5} \Rightarrow 5 r =4 n -4 r +4 \\ & \Rightarrow 9 r =4( n +1) \quad \ldots(1) \end{aligned}$
$\begin{aligned} & \Rightarrow \frac{{ }^n C_r(2)^{ r }}{{ }^{ n } C _{ r +1}(2)^{ r +1}}=\frac{5}{8} \\ & \Rightarrow \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-r-1)!}}=\frac{5}{4} \Rightarrow \frac{r+1}{n-r}=\frac{5}{4} \end{aligned}$
$\Rightarrow 4 r+4=5 n-5 r \Rightarrow 5 n-4=9 r \dots(2)$
From (1) and (2)
$\Rightarrow 4 n +4=5 n -4 \Rightarrow n =8$
(1) $\Rightarrow r=4$
so, coefficient of middle term is
${ }^8 C _4 2^4=16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120$

Hint

Coefficient of $x ^9$ in $\left(\alpha x^3+\frac{1}{\beta x}\right)={ }^{11} C_6 \cdot \frac{\alpha^5}{\beta^6}$
$\because$ Both are equal
$\begin{aligned} & \therefore { }^{11} C_6 \cdot \frac{\alpha^5}{\beta^6}=-{ }^{11} C_6 \cdot \frac{\alpha^6}{\beta^5} \\ & \Rightarrow \frac{1}{\beta}=-\alpha \\ & \Rightarrow \alpha \beta=-1 \\ & \Rightarrow(\alpha \beta)^2=1 \end{aligned}$

Hint

$\begin{aligned} & (2023)^{2023} \\ & =(2030-7)^{2023} \\ & =(35 K -7)^{2023} \end{aligned}$
$={ }^{2023} C _0(35 K )^{2023}(-7)^0+{ }^{2023} C _1(35 K )^{2022}(-7)+\ldots \ldots+\ldots \ldots .+{ }^{2023} C _{2023}(-7)^{2023}$
$=35 N -7^{2023}$
Now,$-7^{2023}=-7 \times 7^{2022}=-7\left(7^2\right)^{1011}$
$=-7(50-1)^{1011}$
$=-7\left({ }^{1011} C _0 50^{1011}-{ }^{1011} C _1(50)^{1010}+\ldots . .{ }^{1011} C _{1011}\right)$
$\begin{aligned} & =-7(5 \lambda-1) \\ & =-35 \lambda+7 \end{aligned}$
$\therefore$ when $(2023)^{2023}$ is divided by 35 remainder is 7.

Hint

$\begin{aligned} & \left(2 x+\frac{1}{x^7}+3 x^2\right)^5 \\ & \frac{1}{x^{35}}\left(2 x^8+1+3 x^9\right)^5 \\ & \frac{1}{x^{35}}\left(1+x^8(3 x+2)\right)^5 \end{aligned}$
Term independent of $x=$ coefficient of $x^{35}$ in
$\begin{aligned} & { }^5 C_4\left(x^8(3 x+2)\right)^4 \\ = & { }^5 C_4 \text { coefficient of } x^3 \text { in }(2+3 x)^4 \\ = & { }^5 C_4 \times{ }^4 C_3(2)^1(3)^3 \\ = & 5 \times 4 \times 2 \times 27 \\ = & 1080 \end{aligned}$

Hint

$\begin{aligned} & S=1-3 n+\frac{9 n(n-1)}{2}=376 \\ & 3 n^2-5 n-250=0 \\ & n=10, \frac{-25}{3} \text { (Rejected) } \\ & T_{r+1}={ }^n C_r \cdot x^{n-r}\left(\frac{-3}{x^2}\right)^r \\ & ={ }^n C_r  x^{n-3 r}(-3)^r \\ & ={ }^{10} C_r x^{10-3 r}(-3)^r \end{aligned}$
Here $r=2$
$\begin{aligned} \text { Required coefficient } & ={ }^{10} C _2(-3)^2 \\ & =45 \times 9 \\ & =405 \end{aligned}$

Hint

$\begin{aligned} & \text { (1) }{ }^n C_r=\frac{n}{r} \cdot{ }^{n-1} C_{r-1} \\ & \text { Given, } \\ & \sum_{r=0}^{2023} r^2 \cdot{ }^{2023} C_r \\ & =\sum_{r=0}^{2023} r^2 \cdot \frac{2023}{r} \cdot{ }^{2022} C_{r-1} \\ & =2023 \sum_{r=0}^{2023} r \cdot{ }^{2022} C_{r-1} \\ & =2023 \sum_{r=0}^{2023}[(r-1)+1] \cdot{ }^{2022} C_{r-1} \\ & =2023\left[\sum_{r=0}^{2023}(r-1) \cdot{ }^{2022} C_{r-1}+\sum_{r=0}^{2023}{ }^{2022} C_{r-1}\right] \\ & =2023\left[\sum_{r=0}^{2023}(r-1) \cdot \frac{2022}{(r-1)} \cdot{ }^{2021} C_{r-2}+2^{2022}\right] \\ & =2023\left[2022 \sum_{r=0}^{2023}{ }^{2021} C_{r-2}+2^{2022}\right] \\ & =2023\left[2022 \cdot 2^{2021}+2^{2022}\right] \\ & =2023 \cdot 2^{2021}[2022+2] \\ & =2023 \cdot 2^{2021} \cdot 2024 \end{aligned}$
$\begin{aligned} & =2023 \cdot \frac{2^{2022}}{2} \cdot 2024 \\ & =2023 \cdot 2^{2022} \cdot 1012 \\ & \therefore \alpha=1012 \end{aligned}$

Hint

Given,
$\begin{aligned} & \sum_{k=1}^{10} k^2\left({ }^{10} C_k\right)^2=2200 L \\ & \Rightarrow \sum_{k=1}^{10}\left(k \cdot{ }^{10} C_k\right)^2=22000 L \\ & \Rightarrow \sum_{k=1}^{10}\left(k \cdot \frac{10}{k} \cdot{ }^9 C_{k-1}\right)^2=22000 L \\ & \Rightarrow \sum_{k=1}^{10}\left(10 \cdot{ }^9 C_{k-1}\right)^2=22000 L \end{aligned}$
$\begin{aligned} & \Rightarrow 100 \cdot \sum_{k=1}^{10}\left({ }^9 C_{k-1}\right)^2=22000 L \\ & \Rightarrow 100\left(\left({ }^9 C_0\right)^2+\left({ }^9 C_1\right)^2+\ldots .+\left({ }^9 C_9\right)^2\right)=22000 L \\ & \Rightarrow 100\left({ }^{18} C_9\right)=22000 L \end{aligned}$
[Note : $\left({ }^n C_1\right)^2+\left({ }^n C_2\right)^2+\ldots+\left({ }^n C_n\right)^2={ }^{2 n} C_n$ ]
$\begin{aligned} & \Rightarrow 100 \times \frac{18!}{9!9!}=22000 L \\ & \Rightarrow L=221 \end{aligned}$

Hint

Fifth term from beginning $={ }^n C_4\left(2^{\frac{1}{4}}\right)^{n-4}\left(3^{\frac{-1}{4}}\right)^4$
Fifth term from end $=(n-5+1)^{\text {th }}$ term from begin $={ }^n C_{n-4}\left(2^{\frac{1}{4}}\right)^3\left(3^{\frac{-1}{4}}\right)^{n-4}$
$\begin{aligned} & \text { Given } \frac{{ }^n C_4 2^{\frac{n-4}{4}} \cdot 3^{-1}}{{ }^n C_{n-3} 2^{\frac{4}{4}} \cdot 3^{-\left(\frac{n-4}{4}\right)}}=6^{\frac{1}{4}} \\ & \Rightarrow 6^{\frac{n-8}{4}}=6^{\frac{1}{4}} \\ & \Rightarrow \frac{n-8}{4}=\frac{1}{4} \Rightarrow n=9 \\ & T_6=T_{5+1}={ }^9 C_5\left(2^{\frac{1}{4}}\right)^4\left(3^{\frac{-1}{4}}\right)^5 \\ & =\frac{{ }^9 C_5 \cdot 2}{3^{\frac{1}{4}} \cdot 3}=\frac{84}{3^{\frac{1}{4}}}=\frac{\alpha}{3^{\frac{1}{4}}} \\ & \Rightarrow \alpha=84 \end{aligned}$

Hint

Coefficients of middle terms of given expansions are ${ }^4 C_2 \frac{1}{6} \beta^2,{ }^2 C_1(-3 \beta),{ }^6 C_3\left(\frac{-\beta}{2}\right)^3$ form an A.P.
$\begin{aligned} & \therefore 2.2(-3 \beta)=\beta^2-\frac{5 \beta^3}{2} \\ & \Rightarrow-24=2 \beta-5 \beta^2 \\ & \Rightarrow 5 \beta^2-2 \beta-24=0 \\ & \Rightarrow 5 \beta^2-12 \beta+10 \beta-24=0 \\ & \Rightarrow \beta(5 \beta-12)+2(5 \beta-12)=0 \\ & \beta=\frac{12}{5} \\ & d=-6 \beta-\beta^2 \\ & \therefore 50-\frac{2 d}{\beta^2}=50-2 \frac{\left(-6 \beta-\beta^2\right)}{\beta^2}=50+\frac{12}{\beta}+2=57 \end{aligned}$

Hint

$l=1+\left(1+{ }^{49} C_0+{ }^{49} C_1+\ldots+{ }^{49} C_{49}\right)\left({ }^{50} C_2+{ }^{50} C_4+\ldots+{ }^{50} C_{50}\right)$
As ${ }^{49} C_0+{ }^{49} C_1+\ldots \ldots+{ }^{49} C_{49}=2^{49}$
and ${ }^{50} C_0+{ }^{50} C_2+\ldots+{ }^{50} C_{50}=2^{49}$
$\begin{aligned} & \Rightarrow{ }^{50} C_2+{ }^{50} C_4+\ldots+{ }^{50} C_{50}=2^{49}-1 \\ & \therefore l=1+\left(2^{49}+1\right)\left(2^{49}-1\right) \\ & =2^{98} \\ & \therefore m=1 \text { and } n=98 \\ & m+n=99 \end{aligned}$

Hint

$(3+6 x)^n=3^n(1+2 x)^n$
If $T _9$ is numerically greatest term
$\begin{aligned} & \therefore T_8 \leq T_9 \leq T_{10} \\ & { }^n C_7 3^{n-7}(6 x)^7 \leq{ }^n C_8 3^{n-8}(6 x)^8 \geq{ }^n C_9 3^{n-9}(6 x)^9 \\ & \Rightarrow \frac{n!}{(n-7)!7!} 9 \leq \frac{n!}{(n-8)!8!} 3 \cdot(6 x) \geq \frac{n!}{(n-9)!9!}(6 x)^2 \\ & \Rightarrow \underbrace{\frac{9}{(n-7)(n-8)}} \leq \underbrace{\frac{18\left(\frac{3}{2}\right)}{(n-8) 8} \geq \frac{36}{9.8} \frac{9}{4}} \end{aligned}$
$\begin{aligned} & 72 \leq 27(n-7) \text { and } 27 \geq 9(n-8) \\ & \frac{29}{3} \leq n \text { and } n \leq 11 \\ & \therefore n_0=10 \\ & \text { For }(3+6 x)^{10} \\ & T_{r+1}={ }^{10} C_r \\ & 3^{10-r}(6 x)^r \end{aligned}$
For coeff. of $x^6$
$r=6 \Rightarrow{ }^{10} C_6 3^4 \cdot 6^6$
For coeff. of $x^3$
$\begin{aligned} & r=3 \Rightarrow{ }^{10} C_3 3^7 \cdot 6^3 \\ & \therefore k=\frac{{ }^{10} C_6}{{ }^{10} C_3} \cdot \frac{3^4 \cdot 6^6}{3^7 \cdot 6^3}=\frac{10!7!3!}{6!4!10!} \cdot 8 \\ & \Rightarrow k=14 \\ & \therefore k+n_0=24 \end{aligned}$

Hint

$\begin{aligned} & \text { Coefficient of } x \text { in }(1+x)^p(1-x)^q \\ & -{ }^p C_0{ }^q C_1+{ }^p C_1{ }^q C_0=-3 \Rightarrow p-q=-3 \\ & \text { Coefficient of } x ^2 \text { in }(1+x)^p(1-x)^q \\ & { }^p C_0{ }^q C_2-{ }^p C_1{ }^q C_1+{ }^p C_2{ }^q C_0=-5 \\ & \frac{q(q-1)}{2}-p q+\frac{p(q-1)}{2}=-5 \\ & \frac{q^2-q}{2}-(q-3) q+\frac{(q-3)(q-4)}{2}=-5 \\ & \Rightarrow q=11, p=8 \\ & \text { Coefficient of } x ^3 \text { in }(1+x)^8(1-x)^{11} \\ & =-{ }^{11} C_3+{ }^8 C_1{ }^{11} C_2-{ }^8 C_2{ }^{11} C_1+{ }^8 C_3=23 \end{aligned}$

Hint

General term of $\left(t^2 x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{t}\right)^{15}$ is
$\begin{aligned} & T_{r+1}={ }^{15} C_r \cdot\left(t^2 x^{\frac{1}{5}}\right)^{15-r} \cdot\left(\frac{(1-x)^{\frac{1}{10}}}{t}\right)^r \\ & ={ }^{15} C_r \cdot t^{30-2 r} \cdot x^{\frac{15-r}{5}} \cdot(1-x)^{\frac{r}{10}} \cdot t^{-r} \\ & ={ }^{15} C_r \cdot t^{30-3 r} \cdot x^{\frac{15-r}{5}} \cdot(1-x)^{\frac{r}{10}} \end{aligned}$
Term will be independent of $t$ when $30-3 r=0 \Rightarrow r=10$
$\therefore T_{10+1}=T_{11}$ will be independent of $t$
$\begin{aligned} & \therefore T_{11}={ }^{15} C_{10} \cdot x^{\frac{15-10}{5}} \cdot(1-x)^{\frac{10}{10}} \\ & ={ }^{15} C_{10} \cdot x^1 \cdot(1-x)^1 \end{aligned}$
$T _{11}$ will be maximum when $x(1-x)$ is maximum.
Let $f(x)=x(1-x)=x-x^2$
$f(x)$ is maximum or minimum when $f^{\prime}(x)=0$
$\therefore f^{\prime}(x)=1-2 x$
For maximum $/$ minimum $f^{\prime}(x)=0$
$\begin{aligned} & \therefore 1-2 x=0 \\ & \Rightarrow x=\frac{1}{2} \end{aligned}$
Now, $f^{\prime \prime}(x)=-2<0$
$\therefore$ At $x=\frac{1}{2}, f(x)$ maximum
$\therefore$ Maximum value of $T _{11}$ is
$\begin{aligned} & ={ }^{15} C_{10} \cdot \frac{1}{2}\left(1-\frac{1}{2}\right) \\ & ={ }^{15} C_{10} \cdot \frac{1}{4} \end{aligned}$
Given $K={ }^{15} C_{10} \cdot \frac{1}{4}$
Now, $8 K=2\left({ }^{15} C_{10}\right)$
$=6006$

Hint

Given, Binomial expansion
$\left(2 x^{\frac{1}{5}}-\frac{1}{x^{\frac{1}{5}}}\right)^{15}$
$\therefore$ General Term
$\begin{aligned} & T_{r+1}={ }^{15} C_r \cdot\left(2 x^{\frac{1}{5}}\right)^{15-r} \cdot\left(-\frac{1}{x^{\frac{1}{5}}}\right)^r \\ & ={ }^{15} C_r \cdot 2^{15-r} \cdot x^{\frac{1}{5}(15-r-r)} \cdot(-1)^r \end{aligned}$
For $x^{-1}$ term;
$\begin{aligned} & \frac{1}{5}(15-2 r)=-1 \\ & \Rightarrow 15-2 r=-5 \\ & \Rightarrow 2 r=20 \\ & \Rightarrow r=10 \end{aligned}$
$m$ is the coefficient of $x^{-1}$ term,
$\begin{aligned} & \therefore m={ }^{15} C_{10} \cdot 2^{15-10} \cdot(-1)^{10} \\ & ={ }^{15} C_{10} \cdot 2^5 \end{aligned}$
For $x^{-3}$ term;
$\begin{aligned} & \frac{1}{5}(15-2 r)=-3 \\ & \Rightarrow 15-2 r=-15 \\ & \Rightarrow 2 r=30 \\ & \Rightarrow r=15 \end{aligned}$
$n$ is the coefficient of $x^{-3}$ term,
$\begin{aligned} & \therefore n={ }^{15} C_{15} \cdot 2^{15-15} \cdot(-1)^{15} \\ & =1 \cdot 1 \cdot-1 \\ & =-1 \end{aligned}$
Given,
$\begin{aligned} & m n^2={ }^{15} C_r \cdot 2^r \\ & \left.\Rightarrow{ }^{15} C_{10} \cdot 2^5 \cdot(1)^2={ }^{15} C_r \cdot 2^r \text { [putting value of } m \text { and } n \right] \\ & \Rightarrow{ }^{15} C_{15-10} \cdot 2^5={ }^{15} C_r \cdot 2^r \\ & \Rightarrow{ }^{15} C_5 \cdot 2^5={ }^{15} C_r \cdot 2^r \end{aligned}$
Comparing both side, we get
$r=5$

Hint

(b) Given Binomial expression is
$\left(2 x^3+\frac{3}{x^k}\right)^{12}$
General term,
$\begin{aligned} & T_{r+1}={ }^{12} C_r\left(2 x^3\right)^r \cdot\left(\frac{3}{x^k}\right)^{12-r} \\ & =\left({ }^{12} C_r \cdot 2^r \cdot 3^{12-r}\right) \cdot x^{3 r-12 k+k r} \end{aligned}$
For constant term,
$\begin{aligned} & 3 r-12 k+k r=0 \\ & \Rightarrow k(12-r)=3 r \\ & \Rightarrow k=\frac{3 r}{12-r} \end{aligned}$
For $r =1, k=\frac{3}{11}$ (not integer)
For $r =2, k=\frac{6}{10}$ (not integer)
For $r =3, k=\frac{9}{9}=1$ (integer)
For $r =6, k=\frac{18}{6}=3$ (integer)
For $r =8, k=\frac{24}{4}=6$ (integer)
For $r =9, k=\frac{27}{3}=9$ (integer)
For $r =10, k=\frac{30}{2}=15$ (integer)
For $r =11, k=\frac{33}{1}=33$ (integer)
So, for $r=3,6,8,9,10$ and 11, $k$ is positive integer.
When $k =1$ then $r =3$ and constant term is
$\begin{aligned} & ={ }^{12} C_3 \cdot 2^3 \cdot 3^9 \\ & =\frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1} \cdot 2^3 \cdot 3^9 \\ & =2 \cdot 11 \cdot 2 \cdot 5 \cdot 2^3 \cdot 3^9 \\ & =11 \cdot 5 \cdot 2^5 \cdot 3^9 \\ & =2^5 \cdot\left(55 \cdot 3^9\right) \\ & =2^5(I) \\ & \neq 2^8 \cdot I \end{aligned}$
When $x =3$ then $r =6$ and constant term
$\begin{aligned} & ={ }^{12} C_6 \cdot 2^6 \cdot 3^6 \\ & =\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \cdot 2^6 \cdot 3^6 \\ & =2^8 \cdot 231 \cdot 3^6 \\ & =2^8(I) \end{aligned}$
When $k =6$ then $r =8$ and constant term
$\begin{aligned} & ={ }^{12} C_8 \cdot 2^8 \cdot 3^4 \\ & =\frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2 \cdot 1} \cdot 2^8 \cdot 3^4 \\ & =2^8 \cdot 55 \cdot 3^6 \\ & =2^8(I) \end{aligned}$
When $x=9$ then $r=9$ and constant term
$\begin{aligned} & ={ }^{12} C_9 \cdot 2^9 \cdot 3^3 \\ & =\frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1} \cdot 2^9 \cdot 3^3 \\ & =2^{11} \cdot 55 \cdot 3^3 \end{aligned}$
Here power of 2 is 11 which is greater than 8 . So, $k=9$ is not possible.
Similarly for $k =15$ and $k =33,2^8 . I$ form is not possible.
$\therefore k =3$ and $k =6$ is accepted.
$\therefore$ For 2 positive integer value of $k , 2^8$. $I$ form of constant term possible.

Hint

Given, Binomial expression is
$\begin{aligned} & =\left(x^n+\frac{2}{x^5}\right)^7 \\ & \therefore \text { General term } \\ & T_{r+1}={ }^7 C_r \cdot\left(x^n\right)^{7-r} \cdot\left(\frac{2}{x^5}\right)^r \\ & ={ }^7 C_r \cdot x^{7 n-n r-5 r} \cdot 2^r \end{aligned}$
For positive power of $x$,
$\begin{aligned} & 7 n-n r-5 r>0 \\ & \Rightarrow 7 n>r(n+5) \\ & \Rightarrow r<\frac{7 n}{n+5} \end{aligned}$
As $r$ represent term of binomial expression so $r$ is always integer.
Given that sum of coefficient is 939 .
When $r=0$,
sum of coefficient $={ }^7 C_0 \cdot 2^0=1$
when $r=1$,
sum of coefficient $={ }^7 C_0 \cdot 2^0+{ }^7 C_1 \cdot 2^1=1+14=15$
when $r=2$,
sum of coefficient
$={ }^7 C_0 \cdot 2^0+{ }^7 C_1 \cdot 2^1+{ }^7 C_2 \cdot 2^2$
$=1+14+84$
$=99$
when $r=3$,
sum of coefficient
$={ }^7 C_0 \cdot 2^0+{ }^7 C_1 \cdot 2^1+{ }^7 C_2 \cdot 2^2+{ }^7 C_3 \cdot 2^3$
$=1+14+84+280$
$=379$
when $r=4$,
sum of coefficient
$\begin{aligned} & ={ }^7 C_0 \cdot 2^0+{ }^7 C_1 \cdot 2^1+{ }^7 C_2 \cdot 2^2+{ }^7 C_3 \cdot 2^3+{ }^7 C_4 \cdot 2^4 \\ & =1+14+84+280+560 \\ & =939 \end{aligned}$
$\therefore$ For $r=4$ sum of coefficient $=939$
To get value of $r=4$, value of $\frac{7 n}{n+5}$ should be between 4 and 5 .
$\begin{aligned} & \therefore 4<\frac{7 n}{n+5}<5 \\ & \Rightarrow 4 n+20<7 n<5 n+25 \\ & \therefore 4 n+20<7 n \\ & \Rightarrow 3 n>20 \\ & \Rightarrow n>\frac{20}{3} \\ & \Rightarrow n>6.66 \end{aligned}$
and
$\begin{aligned} & 7 n<5 n+25 \\ & \Rightarrow 2 n<25 \\ & \Rightarrow n<12.5 \\ & \therefore 6.66<n<12.5 \end{aligned}$
$\therefore$ Possible integer values of $n=7,8,9,10,11,12$
$\therefore$ Sum of values of $n=7+8+9+10+11+12=57$

Hint

Given Binomial Expansion
$=\left(\frac{\sqrt{x}}{5^{\frac{1}{4}}}+\frac{\sqrt{x}}{5^{\frac{1}{3}}}\right)^{60}$
$\therefore$ General term
$\begin{aligned} & T_{r+1}={ }^{60} C_r \cdot\left(\frac{x^{1 / 2}}{5^{1 / 4}}\right)^{60-r} \cdot\left(\frac{5^{1 / 2}}{x^{1 / 3}}\right)^r \\ & ={ }^{60} C_r \cdot 5^{\left(\frac{r}{4}-15+\frac{r}{2}\right)} \cdot x^{\left(30-\frac{r}{2}-\frac{r}{3}\right)} \\ & ={ }^{60} C_r \cdot 5^{\left(\frac{3 r-60}{4}\right)} \cdot x^{\left(\frac{180-5 r}{6}\right)} \end{aligned}$
For $x ^{10}$ term,
$\begin{aligned} & \frac{180-5 r}{6}=10 \\ & \Rightarrow 5 r=120 \\ & \Rightarrow r=24 \end{aligned}$
$\begin{aligned} & \therefore \text { Coefficient of } x^{10}={ }^{60} C_{24} \cdot 5\left(\frac{3 \times 24-60}{4}\right) \\ & ={ }^{60} C_{24} \cdot 5^3 \\ & =\frac{60!}{24!36!} \cdot 5^3 \end{aligned}$
It is given that,
$\frac{60!}{24!36!} \cdot 5^3=5^k \cdot l \dots(1)$
Also given that, I is coprime to 5 means I can’t be multiple of 5 . So we have to find all the factors of 5 in 60 !, 24 ! and 36 !
[Note : Formula for exponent or degree of prime number in n!.
Exponent of $p$ in $n!=\left\lceil\frac{n}{p}\right\rceil+\left\lceil\frac{n}{p^2}\right\rceil+\left\lceil\frac{n}{p^3}\right\rceil+\ldots .$. until 0 comes here $p$ is a prime number. ]
$\therefore$ Exponent of 5 in 60 !
$\begin{aligned} & =\left\lceil\frac{60}{5}\right\rceil+\left\lceil\frac{60}{5^2}\right\rceil+\left\lceil\frac{60}{5^3}\right\rceil+\ldots . . \\ & =12+2+0+\ldots . . \\ & =14 \end{aligned}$
Exponent of 5 in 24 !
$\begin{aligned} & =\left\lceil\frac{24}{5}\right\rceil+\left\lceil\frac{24}{5^2}\right\rceil+\left\lceil\frac{24}{5^3}\right\rceil+\ldots \ldots \\ & =4+0+0 \ldots \ldots \\ & =4 \end{aligned}$
Exponent of 5 in 36 !
$\begin{aligned} & =\left\lceil\frac{36}{5}\right\rceil+\left\lceil\frac{36}{5^2}\right\rceil+\left\lceil\frac{36}{5^3}\right\rceil+\ldots \ldots . . \\ & =7+1+0 \ldots \ldots \\ & =8 \end{aligned}$
$\therefore$ From equation (1), exponent of 5 overall
$\begin{aligned} & \frac{5^{14}}{5^4 \cdot 5^8} \cdot 5^3=5^k \\ & \Rightarrow 5^5=5^k \\ & \Rightarrow k=5 \end{aligned}$

Hint

Here property used is
${ }^n C_r+{ }^n C_{r+1}={ }^{n+1} C_{r+1}$
Given, ${ }^{40} C_0+{ }^{41} C_1+{ }^{42} C_2+\ldots+{ }^{60} C_{20}=\frac{m}{n}{ }^{60} C_{20}$
As ${ }^{40} C_0={ }^{41} C_0=1$
So, we replace ${ }^{40} C_0$ with ${ }^{41} C_0$.
$\begin{aligned} & \Rightarrow{ }^{41} C_0+{ }^{41} C_1+{ }^{42} C_2+\ldots \ldots+{ }^{60} C_{20}=\frac{m}{n} \cdot{ }^{60} C_{20} \\ & \Rightarrow{ }^{42} C_1+{ }^{42} C_2+\ldots \ldots+{ }^{60} C_{20}=\frac{m}{n} \cdot{ }^{60} C_{20} \\ & \Rightarrow{ }^{43} C_2+{ }^{43} C_3+\ldots \ldots+{ }^{60} C_{20}=\frac{m}{n} \cdot{ }^{60} C_{20} \\ & \Rightarrow{ }^{44} C_3+{ }^{44} C_4+\ldots \ldots+{ }^{60} C_{20}=\frac{m}{n} \cdot{ }^{60} C_{20} \\ & \Rightarrow{ }^{45} C_4+{ }^{45} C_5+\ldots \ldots+{ }^{60} C_{20}=\frac{m}{n} \cdot{ }^{60} C_{20} \end{aligned}$
$\vdots$
$\begin{aligned} & \Rightarrow{ }^{60} C_{19}+{ }^{60} C_{20}=\frac{m}{n} \cdot{ }^{60} C_{20} \\ & \Rightarrow{ }^{61} C_{20}=\frac{m}{n} \cdot{ }^{60} C_{20} \\ & \Rightarrow \frac{61!}{20!41!}=\frac{m}{n} \cdot \frac{60!}{20!40!} \\ & \Rightarrow \frac{61}{41}=\frac{m}{n} \\ & \therefore m=61 \text { and } n=41 \\ & \therefore m+n=61+41=102 \end{aligned}$

Hint

Given, Binomial Expansion
$\left(2 x^3+\frac{3}{x}\right)^{10}$
General term
$\begin{aligned} & T_{r+1}={ }^{10} C_r \cdot\left(2 x^3\right)^{10-r} \cdot\left(\frac{3}{x}\right)^r \\ & ={ }^{10} C_r \cdot 2^{10-r} \cdot 3^r \cdot x^{30-3 r} \cdot x^{-r} \\ & ={ }^{10} C_r \cdot 2^{10-r} \cdot 3^r \cdot x^{30-4 r} \end{aligned}$
For positive even power of $x, 30-4 r$ should be even and positive.
For $r =0,30-4 \times 0=30$ (even and positive)
For $r =1,30-4 \times 1=26$ (even and positive)
For $r=2,30-4 \times 2=22$ (even and positive)
For $r=3,30-4 \times 3=18$ (even and positive)
For $r=4,30-4 \times 4=14$ (even and positive)
For $r=5,30-4 \times 5=10$ (even and positive)
For $r=6,30-4 \times 6=6$ (even and positive)
For $r =7,30-4 \times 7=2$ (even and positive)
For $r =8,30-4 \times 8=-2$ (even but not positive)
So, for $r=1,2,3,4,5,6$ and 7 we can get positive even power of $x$.
$\therefore$ Sum of coefficient for positive even power of $x$
$\begin{aligned} & ={ }^{10} C_0 \cdot 2^{10} \cdot 3^0+{ }^{10} C_1 \cdot 2^9 \cdot 3^1+{ }^{10} C_2 \cdot 2^8 \cdot 3^2+{ }^{10} C_3 \cdot 2^7 \cdot 3^3+{ }^{10} C_4 \cdot 2^6 \\ & ={ }^{10} C_{10} \cdot 2^{10} \cdot 3^0+{ }^{10} C_1 \cdot 2^9 \cdot 3^1+\ldots \ldots+{ }^{10} C_{10} \cdot 2^0 \cdot 3^{10}-\left[{ }^{10} C_8 \cdot 2^2 \cdot 3\right. \\ & =(2+3)^{10}-\left[45 \cdot 4 \cdot 3^8+10 \cdot 2 \cdot 3^9+1 \cdot 1 \cdot 3^{10}\right] \end{aligned}$
$\begin{aligned} & =5^{10}-\left[60 \times 3^9+20 \cdot 3^9+3 \cdot 3^9\right] \\ & =5^{10}-(60+20+3) 3^9 \\ & =5^{10}-83 \cdot 3^9 \\ & \therefore \beta=83 \end{aligned}$

Hint

Given,
$C_1+3.2 C_2+5.3 C_3+\ldots . .$. upto 10 terms
$=\frac{\alpha \cdot 2^{11}}{2^\beta-1}\left(C_0+\frac{C_1}{2}+\frac{C_2}{3}+\ldots . . \text { upto } 10 \text { terms }\right)$
Now, L.H.S. :
$\begin{aligned} & C_1+3 \cdot 2 C_2+5 \cdot 3 C_3+\ldots . . \text { upto } 10 \text { terms } \\ & =1 \cdot 1 C_1+3 \cdot 2 C_2+5 \cdot 3 C_3+\ldots . . \text { upto } 10 \text { terms } \\ & =\sum_{r=1}^{10} r \cdot(2 r-1)^{10} C_r \\ & =\sum_{r=1}^{10}\left(2 r^2-r\right) \cdot{ }^{10} C_r \\ & =2 \cdot \sum_{r=1}^{10} r^2 \cdot{ }^{10} C_r-\sum_{r=1}^{10} r \cdot{ }^n C_r \end{aligned}$
[We know, $\sum_{r=1}^n r \cdot{ }^n C_r=n \cdot 2^{n-1}$
$\begin{aligned} & \text { and } \sum_{r=1}^n r^2 \cdot{ }^n C_r=\sum_{r=1}^n\left(r \cdot{ }^n C_r\right) \cdot r \\ & =\sum_{r=1}^n\left(r \cdot \frac{n}{r} \cdot{ }^{n-1} C_{r-1}\right) \cdot r \\ & =\sum_{r=1}^n\left(n \cdot{ }^{n-1} C_{r-1}\right) \cdot r \\ & =n \sum_{r=1}^n(r-1+1)^{n-1} C_{r-1} \\ & =n \cdot \sum_{r=1}^n(r-1) \cdot{ }^{n-1} C_{r-1}+n \cdot \sum_{r=1}^n{ }^{n-1} C_{r-1} \\ & \left.=n \cdot(n-1) \cdot 2^{n-2}+n \cdot 2^{n-1}\right] \\ & =2\left(n(n-1) 2^{n-2}+n \cdot 2^{n-1}\right)-n \cdot 2^{n-1} \end{aligned}$
$\begin{aligned} & \text { Put } n=10 \\ & =2\left(10 \cdot 9 \cdot 2^8+10 \cdot 2^9\right)-10 \cdot 2^9 \\ & =45 \cdot 2^{10}+10 \cdot 2^{10}-5 \cdot 2^{10} \\ & =2^{10}(45+10-5) \\ & =2^{10} \cdot(50) \\ & =25 \cdot 2^{11} \end{aligned}$
R.H.S. :-
$\frac{\alpha \cdot 2^{11}}{2^\beta-1}\left(C_0+\frac{C_1}{2}+\frac{C_2}{3}+\ldots .\right.$. upto 10 terms $)$
$\frac{\alpha \cdot 2^{11}}{2^\beta-1}\left(\frac{C_0}{1}+\frac{C_1}{2}+\frac{C_2}{3}+\ldots .\right.$. upto 10 terms $)$
$\begin{aligned} & \frac{\alpha \cdot 2^{11}}{2^\beta-1}\left(\sum_{r=0}^n \frac{{ }^n C_r}{r+1}\right) \\ & =\frac{\alpha \cdot 2^{11}}{2^\beta-1}\left(\sum_{r=0}^n \frac{{ }^{n+1} C_{r+1}}{n+1}\right) \\ & =\frac{\alpha \cdot 2^{11}}{2^\beta-1}\left(\frac{{ }^{n+1} C_1+{ }^{n+1} C_2+\ldots .+{ }^{n+1} C_{n+1}}{n+1}\right) \\ & =\frac{\alpha \cdot 2^{11}}{2^\beta-1}\left(\frac{{ }^{n+1} C_0+{ }^{n+1} C_2+\ldots .+{ }^{n+1} C_{n+1}-{ }^{n+1} C_0}{n+1}\right) \\ & =\frac{\alpha \cdot 2^{11}}{2^\beta-1}\left(\frac{2^{n+1}-1}{n+1}\right) \end{aligned}$
Putting value of $n=10$, we get
$=\frac{\alpha \cdot 2^{11}}{2^\beta-1}\left(\frac{2^{11}-1}{11}\right)$
Using L.H.S. = R.H.S.
$\begin{aligned} & \Rightarrow 25 \cdot 2^{11}=\frac{\alpha \cdot 2^{11}}{2^\beta-1}\left(\frac{2^{11}-1}{11}\right) \\ & \Rightarrow 25 \cdot 2^{11}=2^{11}\left(\frac{\alpha}{11}\right)\left(\frac{2^{11}-1}{2^\beta-1}\right) \end{aligned}$
By comparing both sides,
$\begin{aligned} & \frac{\alpha}{11}=25 \Rightarrow \alpha=275 \\ & \text { and } \frac{2^{11}-1}{2^\beta-1}=1 \\ & \Rightarrow 2^{11}=2^\beta \\ & \Rightarrow \beta=11 \\ & \therefore \alpha+\beta=275+11=286 \end{aligned}$

Hint

Given,
$\begin{aligned} & 1+3+3^2+3^3+\ldots \ldots+3^{2021} \\ & =3^0+3^1+3^2+3^3+\ldots+3^{2021} \end{aligned}$
This is a G.P with common ratio $=3$
$\begin{aligned} & \therefore \text { Sum }=\frac{1\left(3^{2022}-1\right)}{3-1} \\ & =\frac{3^{2022}-1}{2} \\ & =\frac{\left(3^2\right)^{2011}-1}{2} \\ & =\frac{(10-1)^{1011}-1}{2} \end{aligned}$
$=\frac{\left[{ }^{1011} C_0 \cdot 10^{1011} \_{ }^{1011} C_1 \cdot 10^{1010}+\ldots . . \cdot-{ }^{1011} C_{1009} \cdot(10)^2+{ }^{1011} C_{1010} \cdot 10-{ }^{1011} C_{1011}\right]-1}{2}$
$\begin{aligned} & =\frac{10^2\left[{ }^{1011} C_0 \cdot(10)^{1009} \_{ }^{1011} C_1 \cdot(1008)+\ldots . .{ }^{1011} C_{1000}\right]+10110-1-1}{2} \\ & =\frac{100 k+10110-2}{2} \\ & =\frac{100 k+10108}{2} \\ & =50 k+5054 \\ & =50 k+50 \times 101+4 \\ & =50[k+101]+4 \\ & =50 k^{\prime}+4 \end{aligned}$
$\therefore$ By dividing 50 we get remainder as 4 .

Hint

$\begin{aligned} & (x+y)^n \Rightarrow 2^n=4096 \\ & 2^{10}=1024 \times 2 \\ & \Rightarrow 2^n=2^{12} \\ & 2^{11}=2048 \\ & n=12 \\ & 2^{12}=4096 \\ & { }^{12} C_6=\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \\ & =11 \times 3 \times 4 \times 7 \\ & =924 \end{aligned}$

Hint

$\begin{aligned} & \frac{10!}{\alpha!\beta!\gamma!} a^\alpha(2 b)^\beta \cdot(4 a b)^\gamma \\ & \frac{10!}{\alpha!\beta!\gamma!} a^{\alpha+\gamma} \cdot b^{\beta+\gamma} \cdot 2^\beta \cdot 4^\gamma \end{aligned}$
$\begin{aligned} &\alpha+\beta+\gamma=10 \ldots (1)\\ &\alpha+\gamma=7 \ldots (2) \\ &\beta+\gamma=8 \ldots (3) \end{aligned}$
$(2)+(3)-(1) \Rightarrow \gamma=5$
$\begin{aligned} & \alpha=2 \\ & \beta=3 \\ & \text { so coefficients }=\frac{10!}{2!3!5!} 2^3 .2^{10} \\ & =\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{2 \times 3 \times 2 \times 5!} \times 2^{13} \\ & =315 \times 2^{16} \Rightarrow k=315 \end{aligned}$

Hint

$\begin{aligned} & \left(\frac{x}{4}-\frac{12}{x^2}\right)^{12} \\ & T_{r+1}=(-1)^r \cdot{ }^{12} C_r\left(\frac{x}{4}\right)^{12-r}\left(\frac{12}{x^2}\right)^r \\ & T_{r+1}=(-1)^r \cdot{ }^{12} C_r\left(\frac{1}{4}\right)^{12-r}(12)^r \cdot(x)^{12-3 r} \end{aligned}$
Term independent of $x \Rightarrow 12-3 r=0 \Rightarrow r=4$
$\begin{aligned} & T_5=(-1)^r \cdot{ }^{12} C_r\left(\frac{1}{4}\right)^8(12)^4=\frac{3^6}{4^4} \cdot k \\ & \Rightarrow k =55 \end{aligned}$

Hint

$\begin{aligned} & 3(1+6)^{22}+2 \cdot(1+9)^{22}-44=(3+2-44)+18 \cdot 1 \\ & =-39+18 \cdot 1 \\ & =(54-39)+18(1-3) \\ & =15+18 c_1 \\ & \Rightarrow \text { Remainder }=15 \end{aligned}$

Hint

$\begin{aligned} & A_k=\sum_{i=0}^9{ }^9 C_i{ }^{12} C_{k-i}+\sum_{i=0}^8{ }^8 C_i{ }^{13} C_{k-i} \\ & A_k={ }^{21} C_k+{ }^{21} C_k=2 \cdot{ }^{21} C_k \\ & A_4-A_3=2\left({ }^{21} C_4-{ }^{21} C_3\right)=2(5985-1330) \\ & 190 p=2(5985-1330) \Rightarrow p=49 \end{aligned}$

Hint

$\text { 1. }{ }^n C_0+3 \cdot{ }^n C_1+5 \cdot{ }^n C_2+\ldots+(2 n+1) \cdot{ }^n C_n$
$\begin{aligned} & T_r=(2 r+1)^n C_r \\ & S=\sum T_r \\ & S=\sum(2 r+1)^n C_r=\sum 2 r^n C_r+\sum{ }^n C_r \\ & S=2\left(n .2^{n-1}\right)+2^n=2^n(n+1) \\ & 2^n(n+1)=2^{100} .101 \Rightarrow n=100 \\ & 2\left[\frac{n-1}{2}\right]=2\left[\frac{99}{2}\right]=98 \end{aligned}$

Hint

$\begin{aligned} & { }^n C_7 2^{n-7} \frac{1}{3^7}={ }^n C_8 2^{n-8} \frac{1}{3^8} \\ & \Rightarrow n -7=48 \Rightarrow n =55 \end{aligned}$

Hint

Coeff. of middle term in $(1+ x )^{20}={ }^{20} C_{10}$ & Sum of coeff. of two middle terms in $(1+$ $x )^{19}={ }^{19} C_9+{ }^{19} C_{10}$
$\text { So required ratio }=\frac{{ }^{20} C_{10}}{{ }^{19} C_9+{ }^{19} C_{10}}=\frac{{ }^{20} C_{10}}{{ }^{20} C_{10}}=1$

Hint

$\begin{aligned} & \left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10} \\ & =\left(\frac{\left(x^{1 / 3}\right)^3+\left(1^{1 / 3}\right)^3}{x^{2 / 3}-x^{1 / 3}+1}-\frac{(\sqrt{x})^2-(1)^2}{x-x^{1 / 2}}\right)^{10} \\ & =\left(\frac{\left(x^{1 / 3}+1\right)\left(x^{2 / 3}-x^{1 / 3}+1\right)}{x^{2 / 3}-x^{1 / 3}+1}-\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}\right)^{10} \\ & =\left(\left(x^{1 / 3}+1\right)-\frac{(\sqrt{x}+1)}{\sqrt{x}}\right)^{10} \\ & =\left(\left(x^{1 / 3}+1\right)-\left(1+\frac{1}{\sqrt{x}}\right)\right)^{10} \\ & =\left(x^{1 / 3}-\frac{1}{x^{1 / 2}}\right)^{10} \end{aligned}$
[Note:

For $\left(x^\alpha \pm \frac{1}{x^\beta}\right)^n$ the $(r+1)^{\text {th }}$ term with power $m$ of $x$ is
$\left.r=\frac{n \alpha-m}{\alpha+\beta}\right]$
Here $\alpha=\frac{1}{3}, \beta=\frac{1}{2}$ and $m =0$ then $r=\frac{10 \times \frac{1}{3}-0}{\frac{1}{3}+\frac{1}{2}}=\frac{10}{3} \times \frac{6}{5}=4$
$\therefore T_5$ is the term independent of $x$.
$\therefore T _5={ }^{10} C_4=210$

Hint

$\begin{aligned} & \left(2 x^r+\frac{1}{x^2}\right)^{10} \\ & \text { General term }={ }^{10} C_R\left(2 x^2\right)^{10-R} x^{-2 R} \\ & \Rightarrow 2^{10-R 10} C_R=180 \ldots \ldots .(1) \end{aligned}$
$\begin{aligned} & \&(10- R ) r -2 R =0 \\ & r=\frac{2 R}{10-R} \\ & r=\frac{2(R-10)}{10-R}+\frac{20}{10-R} \\ & \Rightarrow r=-2+\frac{20}{10-R} \dots(2) \end{aligned}$
$R=8$ or 5 reject equation (1) not satisfied
At $R=8$
$\Rightarrow 2^{10-R} \times{ }^{10} C_R=180 \Rightarrow r=8$

Hint

$\begin{aligned} & 11^n>10^n+9^n \\ & \Rightarrow 11^n-9^n>10^n \\ & \Rightarrow(10+1)^n-(10-1)^n>10^n \end{aligned}$
$\Rightarrow 2\left\{{ }^n C_1 \cdot 10^{n-1}+{ }^n C_3 10^{n-10}+{ }^n C_5 10^{n-5}+\ldots \ldots\right\}>10^n$
$\Rightarrow \frac{1}{5}\left[{ }^n C_1 10^n+{ }^n C_3 10^{n-2}+{ }^n C_5 10^{n-4}+\ldots . .\right]>10^n$
$\Rightarrow \frac{1}{5}\left[{ }^n C_1+{ }^n C_3 10^{-2}+{ }^n C_5 10^{-4}+\ldots . .\right]>1$
Clearly the above inequality is true for $n \geq 5$
For $n=4$, we have $\frac{1}{5}\left[4+\frac{4}{10^2}\right]=\frac{4}{5}\left(\frac{101}{100}\right)<1$
$\Rightarrow$ Inequality does not hold good for $n=1,2,3,4$
So, required number of elements $=\{5,6,7$, $\ldots$ $100\}=96$

Hint

$\begin{aligned} & \left(4^{\frac{1}{4}}+5^{\frac{1}{6}}\right)^{120} \\ & T_{r+1}={ }^{120} C_r\left(2^{1 / 2}\right)^{120-r}(5)^{r / 6} \end{aligned}$
for rational terms $r=6 \lambda$
$0 \leq r \leq 120$
So total no of terms are 21.

Hint

$\begin{aligned} & \left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10} \\ & =\left(\frac{\left(x^{1 / 3}\right)^3+\left(1^{1 / 3}\right)^3}{x^{2 / 3}-x^{1 / 3}+1}-\frac{(\sqrt{x})^2-(1)^2}{x-x^{1 / 2}}\right)^{10} \\ & =\left(\frac{\left(x^{1 / 3}+1\right)\left(x^{2 / 3}-x^{1 / 3}+1\right)}{x^{2 / 3}-x^{1 / 3}+1}-\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}\right)^{10} \\ & =\left(\left(x^{1 / 3}+1\right)-\frac{(\sqrt{x}+1)}{\sqrt{x}}\right)^{10} \\ & =\left(\left(x^{1 / 3}+1\right)-\left(1+\frac{1}{\sqrt{x}}\right)\right)^{10} \\ & =\left(x^{1 / 3}-\frac{1}{x^{1 / 2}}\right)^{10}\left(x^{-\frac{1}{2}}\right)^r \end{aligned}$
For being independent of $x: \frac{10-r}{3}-\frac{r}{2}=0 \Rightarrow r=4$
Term independent of $x={ }^{10} C_4=210$

Hint

$\begin{aligned} & \sum_{k=0}^{10}\left(2^2+3 k\right)^{10} C_k \\ & =4 \sum_{k=0}^{10}{ }^{10} C_k+3 \sum_{k=0}^{10} k \cdot{ }^{10} C_k \\ & =4\left(2^{10}\right)+3 \sum_{k=0}^{10} k \cdot \frac{10}{k} \cdot{ }^9 C_{k-1} \\ & =4\left(2^{10}\right)+3 \cdot 10\left(2^9\right) \\ & =4\left(2^{10}\right)+3.5 \cdot 2^{10} \\ & =2^{10}(19) \end{aligned}$
According to question,
$\begin{aligned} & 19\left(2^{10}\right)=\alpha .3^{10}+\beta \cdot 2^{10} \\ & \therefore \alpha=0, \beta=19 \\ & \Rightarrow \alpha+\beta=19 \end{aligned}$

Hint

$\begin{aligned} & T_{r+1}=n_{C_r} x^{n-r} \cdot\left(\frac{a}{x^2}\right)^r \\ & ={ }^n C_r a^r x^{n-3 r} \\ & T_3={ }^n C_2 a^2 x^{n-6}, T_4={ }^n C_3 a^3 x^{n-9}, T_5={ }^n C_4 a^4 x^{n-12} \end{aligned}$
$\Rightarrow a(n-2)=2 \dots(i)$
and $\frac{\text { coefficient of } T_4}{\text { coefficient of } T_5}=\frac{{ }^n C_3 \cdot a^3}{{ }^n C_4 \cdot a^4}=\frac{4}{a(n-3)}=\frac{8}{3}$
$\Rightarrow a(n-3)=\frac{3}{2} \dots(ii)$
by (i) and (ii) $n=6, a=\frac{1}{2}$
for term independent of ‘ $x$ ‘
$\begin{aligned} & n-3 r=0 \Rightarrow r=\frac{n}{3} \Rightarrow r=\frac{6}{3}=2 \\ & T_3={ }^6 C_2\left(\frac{1}{2}\right)^2 x^0=\frac{15}{4}=3.75 \approx 4 \end{aligned}$

Hint

$\begin{aligned} & 2021=17 m-2 \\ & (2021)^{3762}=(17 m-2)^{3762}=\text { multiple of } 17+2^{3762} \\ & =17 \lambda+2^2\left(2^4\right)^{940} \\ & =17 \lambda+4(17-1)^{940} \\ & =17 \lambda+4(17 \mu+1) \\ & =17 k+4 ;(k \in I) \\ & \therefore \text { Remainder }=4 \end{aligned}$

Hint

$A=\sum(-1)^{k_n} C_k\left(\frac{1}{2}\right)^k+\sum(-1)^{k_n} C_k\left(\frac{3}{4}\right)^k+\ldots \ldots$
$\begin{aligned} & =\left(1-\frac{1}{2}\right)^n+\left(1-\frac{3}{4}\right)^n+\ldots \ldots+\left(1-\frac{31}{32}\right)^n \\ & =\left(\frac{1}{2}\right)^n+\left(\frac{1}{2}\right)^{2 n}+\left(\frac{1}{2}\right)^{3 n}+\ldots \ldots+\left(\frac{1}{2}\right)^{5 n} \end{aligned}$
$=\left(\frac{1}{2}\right)^n\left(\frac{1-\left(\frac{1}{2}\right)^{5 n}}{1-\left(\frac{1}{2}\right)^n}\right)=\frac{2^{5 n}-1}{2^{5 n}\left(2^n-1\right)}$
$\therefore 63 A=\frac{63\left(2^{5 n}-1\right)}{2^{5 n}\left(2^n-1\right)}=\frac{63}{\left(2^n-1\right)}\left(1-\frac{1}{2^{5 n}}\right)$
Given, $63 A=1-\frac{1}{2^{30}}$
$\therefore \frac{63}{\left(2^n-1\right)}\left(1-\frac{1}{2^{5 n}}\right)=1-\frac{1}{2^{30}}$
For $n=6$, L.H.S $=$ R.H.S
$\therefore n =6$

Hint

$30\left({ }^{30} C_0\right)+29\left({ }^{30} C_1\right)+\ldots .+2\left({ }^{30} C_{28}\right)+1\left({ }^{30} C_{29}\right)$
$=30\left({ }^{30} C_{30}\right)+29\left({ }^{30} C_{29}\right)+\ldots \ldots+2\left({ }^{30} C_2\right)+1\left({ }^{30} C_1\right)$
$\begin{aligned} & =\sum_{r=1}^{30} r\left({ }^{30} C_r\right) \\ & =\sum_{r=1}^{30} r\left(\frac{30}{r}\right)\left({ }^{29} C_{r-1}\right) \\ & =30 \sum_{r=1}^{30}{ }^{29} C_{r-1} \end{aligned}$
$\begin{aligned} & =30\left({ }^{29} C_0+{ }^{29} C_1+{ }^{29} C_2+\ldots .+{ }^{29} C_{29}\right) \\ & =30\left(2^{29}\right)=15(2)^{30}=n(2)^m \end{aligned}$
$\begin{aligned} & \therefore n =15, m =30 \\ & \Rightarrow n + m =45 \end{aligned}$

Hint

(d)

$\begin{aligned} & \text { Let } x=4 k+3 \\ & (2020+x)^{2022} \\ & =(2020+4 k+3)^{2022} \\ & =(4(505)+4 k+3)^{2022} \\ & =(4 P+3)^{2022} \\ & =(4 P+4-1)^{2022} \\ & =(4 A-1)^{2022} \\ & 2022 C_0(4 A)^0(-1)^{2022}+{ }^{2022} C_1(4 A)^1(-1)^{2021}+\ldots . . \\ & =1+2022(4 A)(-1)+\ldots . . \\ & =1+8 \lambda \end{aligned}$
$\therefore$ Remainder is 1 .

Hint

$\begin{aligned} & \because 7^n=(10-3)^n=10 k+(-3)^n \\ & 7^n+3^n=10 k+(-3)^n+3^n \end{aligned}$
$10 K+(-3)^n+3^n$
$10 K$ if $n=$ odd $10 K+2.3^n$ if $n=$ even
$\text { Let } n =2 t ; t \in N$
$\begin{aligned} & \therefore 3^n=3^{2 t}=(10-1)^t \\ & =10 p+(-1)^t \\ & =10 p \pm 1 \end{aligned}$
$\therefore$ if $n=$ even then $7^n+3^n$ will not be multiply of 10

So if $n$ is odd then only $7^n+3^n$ will be multiply of 10
$\therefore n=11,13,15 \ldots 99$
Hence, there are 45 two digit number.

Hint

Finding the maximum value for which the sum
$\sum_{i=0}^k\binom{10}{i}\binom{15}{k-i}+\sum_{i=0}^{k+1}\binom{12}{i}\binom{13}{k+1-i}:$
Given that $\binom{n}{r}= \begin{cases}n C_r, & \text { if } n \geq r \geq 0 \\ 0, & \text { Otherwise }\end{cases}$
$\begin{aligned} & (1+x)^{10}=10 C_0+10 C_1 x+10 C_2 x^2+\ldots \ldots+10 C_{10} x^{10} \\ & (1+x)^{15}=15 C_0+15 C_1 x+15 C_2 x^2+\ldots \ldots+15 C_{k-1} x^{k-1}+15 C_{k+1} \\ & x^{k+1}+\ldots 15 C_{15} x^{15} \\ & \sum_{i=0}^k\left(10 C_i\right)\left(15 C_{k-i}\right)=10 C_0 15 C_k+10 C_1 15 C_{k-1}+\ldots+10 C_k 15 C_0 \end{aligned}$
Coefficient of $x_k$ in $(1+x)^{25}=25 C_k$
$\sum_{i=0}^{k+1}\left(12 C_i\right)\left(13 C_{k+1-i}\right)=12 C_0 13 C_{k+1}+12 C_1 13 C_k+\ldots+12 C_{k+1} 13 C_0$
Coefficient of $x^{k+1}$ in $(1+x)^{25}=25 C_{k+1}$
$25 C_k+25 C_{k+1}=26 C_{k+1}$
For maximum value
As per given, $k$ can be as large as possible.
Hence, $k$ can be as large as possible .

Hint

$\begin{aligned} & T_{r+1}={ }^9 C_r\left(\frac{x^{-2 / 5}}{2}\right)^r\left(x^{2 / 3}\right)^{9-r} \\ & ={ }^9 C_r \frac{1}{2^r} x^{\frac{2}{3}(9-r)+\left(\frac{-2 r}{5}\right)} \\ & ={ }^9 C_r \cdot \frac{1}{2^r} \cdot x^{6-\frac{16 r}{15}} \end{aligned}$
For coefficient of $x^{2 / 3}$
$\begin{aligned} & \Rightarrow 6-\frac{16 r}{15}=\frac{2}{3} \\ & \Rightarrow 90-16 r=10 \\ & \Rightarrow r=5 \end{aligned}$
For coefficient of $x^{-2 / 5}$
$\begin{aligned} & \Rightarrow 6-\frac{16 r}{15}=\frac{-2}{5} \\ & \Rightarrow 90-16 r=-6 \\ & \Rightarrow r=6 \end{aligned}$
Sum of coefficient of $x^{2 / 3} \& x^{-2 / 5}$
$\begin{aligned} & ={ }^9 C_5 \cdot \frac{1}{2^5}+{ }^9 C_6 \cdot \frac{1}{2^6} \\ & =\frac{9!}{5!4!}\left(\frac{1}{2^5}\right)+\frac{9!}{6!3!}\left(\frac{1}{2^6}\right)=\frac{21}{4} \end{aligned}$

Hint

$x^2(1+x)^{98}+x^3(1+x)^{97}+\ldots+x^{54}(1+x)^{46}$
It is a G.P. with first term $=x^2(1+x)^{98}$ and common ratio $=\frac{x}{1+x}$
sum of these term $=x^2(1+x)^{98}\left(\frac{\left(\frac{x}{1+x}\right)^{53}-1}{\frac{x}{1+x}-1}\right)$
$=x^2(1+x)^{98}\left((1+x)-x^{53}(1+x)^{-52}\right)$
$=x^2(\underbrace{1+x)^{99}}_{\begin{array}{l} \text { Coeff } \\ \text { of } x^{68} \end{array}}-x^{55} \underbrace{(1+x)^{46}}_{\text {Coeff of } x^{15}}$
$\begin{aligned} & ={ }^{99} C _{68}-{ }^{46} C _{15} \\ & \Rightarrow p=68, q=15 \\ & \Rightarrow p+q=83 \end{aligned}$

Hint

$\begin{aligned} & \left(\sqrt{a} x^2+\frac{1}{2 x^3}\right)^{10} \\ & T_{r+1}={ }^{10} C_r\left(\sqrt{a} x^2\right)^{10-r}\left(\frac{1}{2 x^3}\right)^r \end{aligned}$
Independent of $x \Rightarrow 20-2 r-3 r=0$
$r=4$
Independent of $x$ is ${ }^{10} C_4(\sqrt{a})^6\left(\frac{1}{2}\right)^4=105$
$\begin{gathered} \frac{210}{2 \times 8} a^3=105 \\ \Rightarrow \quad a=2 \\ a^2=4 \end{gathered}$