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

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

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 \(\left.\left({ }^6 C_0(1+2 x)^6\left(3 x^2\right)^0\right)\right]+[\) Cofficient of \(x^2\) in \(\left.\left({ }^6 C_1(1+2 x)^5\left(3 x^2\right)^1\right)\right]\)
\(-\left[{ }^6 \mathrm{C}_1\left(4 x^2\right)\right]\)
\(=60+6 \times 3-24=54\)
\(\therefore\) 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 \(y=\sum_{t=1}^{10} A_r\left(B_{10} B_r-C_{10} A_r\right)\).
Now, \(\sum_{t=1}^{10} A_f B_r=\) coefficient of \(\left[(1+x)^{10}(1+x)^{20}\right]-1\)
\(
\Rightarrow \sum_{\mathrm{r}=1}^{10} \mathrm{~A}_{\mathrm{r}} \mathrm{B}_{\mathrm{r}}=\mathrm{C}_{20}-1=\mathrm{C}_{10}-1
\)
And \(\sum_{t=1}^{10} A_r^2=\) 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 \((a-b)^n, \mathrm{n} \geq 5\);
\(
\begin{aligned}
& T_5+T_6=0 \\
& \Rightarrow{ }^n C_4 a^{n-4} b^4-{ }^n C_5 a a^{n-5} b^5=0 \\
& \Rightarrow \frac{{ }^n C_4}{{ }^n C_5} \cdot \frac{a}{b}=1 \Rightarrow \frac{5}{n-4} \cdot \frac{a}{b}=1 \Rightarrow \quad \frac{a}{b}=\frac{n-4}{5}
\end{aligned}
\)

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 \(=[latex] Coeff. of [latex](\mathrm{r}+2)^{\mathrm{th}}\) 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 ism \(=16\) 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 existance 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.
\(={ }^4 C_0 \times{ }^7 C_1 \times{ }^{12} C_2+{ }^4 C_1 \times{ }^7 C_3 \times{ }^{12} C_0\)
[Coeff. of \(x^c\) in \(\left.(1+x)^4\right]\)
\(
=462+140+504+7=1113 \quad+{ }^4 C_2 \times{ }^7 C_1 \times{ }^{12} C_1+{ }^4 C_4 \times{ }^7 C_1 \times{ }^{12} C_0
\)

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) \(\begin{aligned} \sum_{i=0}^m{ }^{m 0} C_i^{20} C_{m-i}={ }^{10} C_0{ }^{20} C_m & +{ }^{10} C_1{ }^{20} C_{m-1} \\ & +{ }^{10} C_2{ }^{20} C_{m-2}+\ldots . .+{ }^{10} C_m{ }^{20} C_0\end{aligned}\) \(=\) 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

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

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

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

Hint

Coefficient of \(x^{30}\) in \(\frac{(1+x)^6\left(1+x^2\right)^7\left(1-x^3\right)^8}{x^6}\)
\(\Rightarrow\) Coefficient of \(x^{36}\) in \((1+x)^6\left(1+x^2\right)^7\left(1-x^3\right)^8\)
\(\Rightarrow\) General term \(={ }^6 C_{r_1}{ }^7 C_{r_2}{ }^8 C_{r_3}(-1)^{r_3} x^{r_1+2 r_2+3 r_3}\)
\(\Rightarrow r_1+2 r_2+3 r_3=36\)
Case-I:
\(
\begin{tabular}{|c|c|c|}
\hline [latex]r _1[latex] & [latex]r _2[latex] & [latex]r _3[latex] \\
\hline 0 & 6 & 8 \\
\hline 2 & 5 & 8 \\
\hline 4 & 4 & 8 \\
\hline 6 & 3 & 8 \\
\hline
\end{tabular}
\)
\(
r_1+2 r_2=12 \quad\left(\text { Taking } r_3=8\right)
\)
Case-II :
\(
\begin{tabular}{|l|l|l|}
\hline[latex]r_1[latex] & [latex]r_2[latex] & [latex]r_3[latex] \\
\hline 1 & 7 & 7 \\
\hline 3 & 6 & 7 \\
\hline 5 & 5 & 7 \\
\hline
\end{tabular}
\)
\(
r_1+2 r_2=15\left(\text { Taking } r_3=7\right)
\)
\(
\begin{aligned}
&\text { Case-III : }\\
&\begin{array}{|l|l|l|}
\hline r_1 & r_2 & r_3 \\
\hline 4 & 7 & 6 \\
\hline 6 & 6 & 6 \\
\hline
\end{array}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Coefficient }=7+(15 \times 21)+(15 \times 35)+(35) \\
& -(6 \times 8)-(20 \times 7 \times 8)-(6 \times 21 \times 8)+(15 \times 28) \\
& +(7 \times 28)=-678=\alpha \\
& |\alpha|=678
\end{aligned}
\)

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
\(
\begin{aligned}
& \therefore \frac{11}{C_6} \cdot \frac{\alpha^5}{\beta^6}=-\frac{11}{C_5} \cdot \frac{\alpha^6}{\beta^5} \\
& \Rightarrow \frac{1}{\beta}=-\alpha \\
& \Rightarrow \alpha \beta=-1 \\
& \Rightarrow(\alpha \beta)^2=1
\end{aligned}
\)

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

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

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.