NCERT Exemplar MCQs

Hint

We have \(\mathrm{T}_{\mathrm{r}}={ }^{2 r} \mathrm{C}_{r-1}(x)^{2 r-r+1}\left(\frac{1}{x}\right)^{2 r}\).
\(
\begin{aligned}
& =\frac{2 r!}{(r-1)!(r+1)!} x^{r+1-r+1} \\
& =\frac{2 r!}{(r-1)!(r+1)!} x^2
\end{aligned}
\)

Hint

Put \(1-x=y\). Then
\(
\begin{aligned}
\left(1-x+x^2\right)^4= & \left(y+x^2\right)^4 \\
= & { }^4 \mathrm{C}_0 \quad y^4\left(x^2\right)^0+{ }^4 \mathrm{C}_1 y^3\left(x^2\right)^1 \\
& +{ }^4 \mathrm{C}_2 \quad y^2\left(x^2\right)^2+{ }^4 \mathrm{C}_3 \quad y\left(x^2\right)^3+{ }^4 \mathrm{C}_4\left(x^2\right)^4 \\
= & y^4+4 y^3 x^2+6 y^2 x^4+4 y x^6+x^8 \\
= & (1-x)^4+4 x^2(1-x)^3+6 x^4(1-x)^2+4 x^6(1-x)+x^8 \\
= & 1-4 x+10 x^2-16 x^3+19 x^4-16 x^5+10 x^6-4 x^7+x^8
\end{aligned}
\)

Hint

Since \(r^{\text {th }}\) term from the end in the expansion of \((a+b)^n\) is \((n-r+2)^{\text {th }}\) term from the beginning. Therefore \(4^{\text {th }}\) term from the end is \(9-4+2\), i.e., \(7^{\text {th }}\) term from the beginning, which is given by
\(
\mathrm{T}_7={ }^9 \mathrm{C}_6\left(\frac{x^3}{2}\right)^3\left(\frac{-2}{x^2}\right)^6={ }^9 \mathrm{C}_3 \frac{x^9}{8} \cdot \frac{64}{x^{12}}=\frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times \frac{64}{x^3}=\frac{672}{x^3}
\)

Hint

Putting \(\sqrt{1-x^2}=y\), we get
The given expression \(=\left(x^2-y\right)^4+\left(x^2+y\right)^4=2\left[x^8+{ }^4 \mathrm{C}_2 x^4 y^2+{ }^4 \mathrm{C}_4 y^4\right]\)
\(
\begin{aligned}
& =2 x^8+\frac{4 \times 3}{2 \times 1} x^4 \cdot\left(1-x^2\right)+\left(1-x^2\right)^2 \\
& =2\left[x^8+6 x^4\left(1-x^2\right)+\left(1-2 x^2+x^4\right]\right. \\
& =2 x^8-12 x^6+14 x^4-4 x^2+2
\end{aligned}
\)

Hint

Let the general term, i.e., \((r+1)^{\text {th }}\) contain \(x^{11}\).
We have
\(
\begin{aligned}
\mathrm{T}_{r+1} & ={ }^{12} \mathrm{C}_r\left(x^3\right)^{12-r}\left(-\frac{2}{x^2}\right)^r \\
& ={ }^{12} \mathrm{C}_r x^{36-3 r-2 r}(-1)^r 2^r \\
& ={ }^{12} \mathrm{C}_r(-1)^r 2^r x^{36-5 r}
\end{aligned}
\)
Now for this to contain \(x^{11}\), we observe that
\(
36-5 r=11 \text {, i.e., } r=5
\)
Thus, the coefficient of \(x^{11}\) is
\(
{ }^{12} C_5(-1)^5 2^5=-\frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2} \times 32=-25344
\)

Hint

Let \(\mathrm{T}_{r+1}\) contain \(x^{10}\). Then
\(
\begin{aligned}
\mathrm{T}_{r+1} & ={ }^{18} \mathrm{C}_r\left(x^2\right)^{18-r} \frac{-2}{x} \\
& ={ }^{18} \mathrm{C}_r x^{36-2 r}(-1)^r \cdot 2^r x^r \\
& =(-1)^r 2^r{ }^{18} \mathrm{C}_r x^{36-3 r}
\end{aligned}
\)
Thus,
\(
36-3 r=10 \text {, i.e., } r=\frac{26}{3}
\)

\(
\text { Since } r \text { is a fraction, the given expansion cannot have a term containing } x^{10} \text {. }
\)

Hint

Let \((r+1)^{\text {th }}\) term be independent of \(x\) which is given by
\(
\mathrm{T}_{r+1}={ }^{10} \mathrm{C}_r \sqrt{\frac{x}{3}}^{10-r}{\frac{\sqrt{3}}{2 x^2}}^r
\)
\(
\begin{aligned}
& ={ }^{10} \mathrm{C}_r \frac{x}{3}{ }^{\frac{10-r}{2}} 3^{\frac{r}{2}} \frac{1}{2^r x^{2 r}} \\
& ={ }^{10} \mathrm{C}_r 3^{\frac{r}{2}-\frac{10-r}{2}} 2^{-r} x^{\frac{10-r}{2}-2 r} \\
&
\end{aligned}
\)
Since the term is independent of \(x\), we have
\(
\frac{10-r}{2}-2 r=0 \quad \Rightarrow \quad r=2
\)
Hence \(3^{\text {rd }}\) term is independent of \(x\) and its value is given by
\(
\mathrm{T}_3={ }^{10} \mathrm{C}_2 \frac{3^{-3}}{4}=\frac{10 \times 9}{2 \times 1} \times \frac{1}{9 \times 12}=\frac{5}{12}
\)

Hint

Since the power of binomial is even, it has one middle term which is the \(\frac{12+2}{2}\) th term and it is given by
\(
\begin{aligned}
\mathrm{T}_7 & ={ }^{12} \mathrm{C}_6(2 a x)^6\left(\frac{-b}{x^2}\right)^6 \\
& ={ }^{12} \mathrm{C}_6 \frac{2^6 a^6 x^6 \cdot(-b)^6}{x^{12}} \\
& ={ }^{12} \mathrm{C}_6 \frac{2^6 a^6 b^6}{x^6}=\frac{59136 a^6 b^6}{x^6}
\end{aligned}
\)

Hint

Since the power of binomial is odd. Therefore, we have two middle terms which are \(5^{\text {th }}\) and \(6^{\text {th }}\) terms. These are given by
\(
\mathrm{T}_5={ }^9 \mathrm{C}_4\left(\frac{p}{x}\right)^5\left(\frac{x}{p}\right)^4={ }^9 \mathrm{C}_4 \frac{p}{x}=\frac{126 p}{x}
\)
and
\(
\mathrm{T}_6={ }^9 \mathrm{C}_5\left(\frac{p}{x}\right)^4\left(\frac{x}{p}\right)^5={ }^9 \mathrm{C}_5 \frac{x}{p}=\frac{126 x}{p}
\)

Hint

We have
\(
\begin{aligned}
2^{4 n+4}-15 n-16= & 2^{4(n+1)}-15 n-16 \\
= & 16^{n+1}-15 n-16 \\
= & (1+15)^{n+1}-15 n-16 \\
= & { }^{n+1} \mathrm{C}_0 15^0+{ }^{n+1} \mathrm{C}_1 15^1+{ }^{n+1} \mathrm{C}_2 15^2+{ }^{n+1} \mathrm{C}_3 15^3 \\
& +\ldots+{ }^{n+1} \mathrm{C}_{n+1}(15)^{n+1}-15 n-16 \\
= & 1+(n+1) 15+{ }^{n+1} \mathrm{C}_2 15^2+{ }^{n+1} \mathrm{C}_3 15^3 \\
& +\ldots+{ }^{n+1} \mathrm{C}_{n+1}(15)^{n+1}-15 n-16 \\
= & 1+15 n+15+{ }^{n+1} \mathrm{C}_2 15^2+{ }^{n+1} \mathrm{C}_3 15^3 \\
& +\ldots+{ }^{n+1} \mathrm{C}_{n+1}(15)^{n+1}-15 n-16 \\
= & 15^2\left[{ }^{n+1} \mathrm{C}_2+{ }^{n+1} \mathrm{C}_3 15+\ldots \text { so on }\right]
\end{aligned}
\)
Thus, \(2^{4 n+4}-15 n-16\) is divisible by 225 .

Hint

\(
\text { We have }(2+3 x)^9=2^9\left(1+\frac{3 x}{2}\right)^9
\)
\(
\frac{\mathrm{T}_{r+1}}{\mathrm{~T}_r}=\frac{2^9\left[{ }^9 \mathrm{C}_r\left(\frac{3 x}{2}\right)^r\right]}{2^9\left[{ }^9 \mathrm{C}_{r-1}\left(\frac{3 x}{2}\right)^{r-1}\right]}
\)
\(
=\frac{{ }^9 \mathrm{C}_r}{{ }^9 \mathrm{C}_{r-1}}\left|\frac{3 x}{2}\right|
\)
\(
=\frac{10-r}{r}\left|\frac{3 x}{2}\right|=\frac{10-r}{r}\left(\frac{9}{4}\right) \quad \text { Since } \quad x=\frac{3}{2}
\)
Therefore,
\(
\begin{aligned}
\frac{\mathrm{T}_{r+1}}{\mathrm{~T}_r} \geq 1 & \Rightarrow \frac{90-9 r}{4 r} \geq 1 \\
& \Rightarrow 90-9 r \geq 4 r \\
& \Rightarrow r \leq \frac{90}{13} \\
& \Rightarrow r \leq 6 \frac{12}{13}
\end{aligned}
\)
Thus the maximum value of \(r\) is 6 . Therefore, the greatest term is \(\mathrm{T}_{r+1}=\mathrm{T}_7\).
Hence,
\(
\begin{aligned}
\mathrm{T}_7 & =2^9\left[{ }^9 \mathrm{C}_6\left(\frac{3 x}{2}\right)^6\right], \quad \text { where } x=\frac{3}{2} \\
& =2^9 \cdot{ }^9 \mathrm{C}_6\left(\frac{9}{4}\right)^6=2^9 \cdot \frac{9 \times 8 \times 7}{3 \times 2 \times 1}\left(\frac{3^{12}}{2^{12}}\right)=\frac{7 \times 3^{13}}{2}
\end{aligned}
\)

Hint

We have
\(
(1+x)^n 1+\frac{1}{x}^n=(1+x)^n \quad \frac{x+1}{x} \quad=\frac{(1+x)^{2 n}}{x^n}
\)
Now to find the coefficient of \(x^{-1}\) in \((1+x)^n \quad 1+\frac{1}{x}^n\), it is equivalent to finding coefficient of \(x^{-1}\) in \(\frac{(1+x)^{2 n}}{x^n}\) which in turn is equal to the coefficient of \(x^{n-1}\) in the expansion of \((1+x)^{2 n}\).
Since \((1+x)^{2 n}={ }^{2 n} \mathrm{C}_0 x^0+{ }^{2 n} \mathrm{C}_1 x^1+{ }^{2 n} \mathrm{C}_2 x^2+\ldots+{ }^{2 n} \mathrm{C}_{n-1} x^{n-1}+\ldots+{ }^{2 n} \mathrm{C}_{2 n} x^{2 n}\)
Thus the coefficient of \(x^{n-1}\) is \({ }^{2 n} \mathrm{C}_{n-1}\)

Hint

We have \((101)^{50}=(100+1)^{50}\)
\(
=100^{50}+50(100)^{49}+\frac{50.49}{2.1}(100)^{48}+\frac{50.49 .48}{3.2 .1}(100)^{47}+\ldots \dots(1)
\)
Similarly \(\quad 99^{50}=(100-1)^{50}\)
\(
=100^{50}-50.100^{49}+\frac{50.49}{2.1}(100)^{48}-\frac{50.49 .48}{3.2 .1}(100)^{47}+\ldots \dots(2)
\)
Subtracting (2) from (1), we get
\(
\begin{aligned}
& 101^{50}-99^{50}=2 \\
& 50 \cdot(100)^{49}+\frac{50 \cdot 49 \cdot 48}{3 \cdot 2 \cdot 1} 100^{47}+\ldots \\
& \Rightarrow \quad 101^{50}-99^{50}=100^{50}+2 \frac{50 \cdot 49 \cdot 48}{3 \cdot 2 \cdot 1} 100^{47}+\ldots \\
& \Rightarrow \quad 101^{50}-99^{50}>100^{50} \\
& \text { Hence } 101^{50}>99^{50}+100^{50}
\end{aligned}
\)

Hint

Since the above series is a geometric series with the common ratio \(\frac{x}{1+x}\)
\(
a=(1+x)^{1000}
\)
\(
r=\frac{x}{x+1}
\)
\(
\mathrm{n}=1001
\)
its sum is
\(
\frac{(1+x)^{1000} 1-\frac{x}{1+x}}{1-\frac{x}{1+x}}
\)
\(
=\frac{(1+x)^{1000}-\frac{x^{1001}}{1+x}}{\frac{1+x-x}{1+x}}=(1+x)^{1001}-x^{1001}
\)
\(
\text { Hence, coefficient of } x^{50} \text { is given by }
\)
\(
{ }^{1001} \mathrm{C}_{50} = \frac{(1001) !}{(951) !(50) !}
\)

Hint

Let \(a_1, a_2, a_3\) and \(a_4\) be the coefficient of four consecutive terms \(\mathrm{T}_{r+1}, \mathrm{~T}_{r+}\) \({ }_2, \mathrm{~T}_{r+3}\), and \(\mathrm{T}_{r+4}\) respectively. Then
\(
\begin{aligned}
& a_1=\text { coefficient of } \mathrm{T}_{r+1}={ }^n \mathrm{C}_r \\
& a_2=\text { coefficient of } \mathrm{T}_{r+2}={ }^n \mathrm{C}_{r+1} \\
& a_3=\text { coefficient of } \mathrm{T}_{r+3}={ }^n \mathrm{C}_{r+2} \\
& a_4=\text { coefficient of } \mathrm{T}_{r+4}={ }^n \mathrm{C}_{r+3}
\end{aligned}
\)
and, Thus
\(
\begin{aligned}
\frac{a_1}{a_1+a_2} & =\frac{{ }^n \mathrm{C}_r}{{ }^n \mathrm{C}_r+{ }^n \mathrm{C}_{r+1}} \\
& =\frac{{ }^n \mathrm{C}_r}{{ }^{n+1} \mathrm{C}_{r+1}} \quad\left(\because{ }^n \mathrm{C}_r+{ }^n \mathrm{C}_{r+1}={ }^{n+1} \mathrm{C}_{r+1}\right) \\
& =\frac{r+1}{n+1}
\end{aligned}
\)
Similarly,
\(
\begin{aligned}
\frac{a_3}{a_3+a_4} & =\frac{{ }^n \mathrm{C}_{r+2}}{{ }^n \mathrm{C}_{r+2}+{ }^n \mathrm{C}_{r+3}} \\
& =\frac{{ }^n \mathrm{C}_{r+2}}{{ }^{n+1} \mathrm{C}_{r+3}}=\frac{r+3}{n+1}
\end{aligned}
\)
Hence,
\(
\text { L.H.S. }=\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{r+1}{n+1}+\frac{r+3}{n+1}=\frac{2 r+4}{n+1}
\)
and
\(
\text { R.H.S. }=\frac{2 a_2}{a_2+a_3}=\frac{2\left({ }^n \mathrm{C}_{r+1}\right)}{{ }^n \mathrm{C}_{r+1}+{ }^n \mathrm{C}_{r+2}}=\frac{2\left({ }^n \mathrm{C}_{r+1}\right)}{{ }^{n+1} \mathrm{C}_{r+2}}
\)
\(
=\frac{2(r+2)}{n+1}
\)

Hint

(c)

\(
\begin{array}{l}
\text { for }(x+a)^{51}-(x-a)^{51} \\
\text { Since }(x+a)^{51}={ }^{51} C_0 x^{51}+{ }^{51} C_1 x^{50} a+{ }^{51} C_2 x^{49} a^2+\ldots+{ }^{51} C_{51} a^{51} \\
\text { and }(x-a)^{51}={ }^{51} C_0 x^{51}-{ }^{51} C_1 x^{50} a+{ }^{51} C_2 x^{49} a^2 \ldots{ }^{51} C_{51} a^{51}
\end{array}
\)
Subtracting above values,
\(
(x+a)^{51}-(x-a)^{51}=2\left({ }^{51} C_1 x^{50} a+{ }^{51} C_3 x^{48} a^3+\ldots+{ }^{51} C_{51} a^{51}\right)
\)
i.e. \(1^{\text {st, }} 3^{\text {rd }}, 5^{\text {th, }} 7^{\text {th }} \ldots\) \(49^{\text {th }}, 51^{\text {th }}\) term are there.
\(\therefore\) applying A.P.
\(
a+(n-1) d=51
\)
i.e. \(1+(n-1) 2=51\)
i.e. \(2(n-1)=50\)
i.e. \(n=26\)

Hint

b is the correct choice. Since \(\mathrm{T}_{r+1}={ }^n \mathrm{C}_r a^{n-r} x^r\) in expansion of \((a+x)^n\),
Therefore,
\(
\mathrm{T}_8={ }^n \mathrm{C}_7(2)^{n-7}\left(\frac{x}{3}\right)^7={ }^n \mathrm{C}_7 \frac{2^{n-7}}{3^7} x^7
\)
and
\(
\mathrm{T}_9={ }^n \mathrm{C}_8(2)^{n-8}\left(\frac{x}{3}\right)^8={ }^n \mathrm{C}_8 \frac{2^{n-8}}{3^8} x^8
\)
Therefore, \({ }^n \mathrm{C}_7 \frac{2^{n-7}}{3^7}={ }^n \mathrm{C}_8 \frac{2^{n-8}}{3^8}\) (since it is given that coefficient of \(x^7=\operatorname{coefficient} x^8\) )
\(
=\frac{2^{n-8}}{3^8} \cdot \frac{3^7}{2^{n-7}}
\)
\(
\Rightarrow \quad \frac{8}{n-7}=\frac{1}{6} \Rightarrow n=55
\)

Hint

a is the correct choice. Putting \(x=1\) and \(-1\) in
we get
\(
\begin{aligned}
\left(1-x+x^2\right)^n & =a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n} \\
1 & =a_0+a_1+a_2+a_3+\ldots+a_{2 n} \dots(1)\\
3^n & =a_0-a_1+a_2-a_3+\ldots+a_{2 n} \dots(2)
\end{aligned}
\)
and
Adding (1) and (2), we get
\(
3^n+1=2\left(a_0+a_2+a_4+\ldots+a_{2 n}\right)
\)
Therefore \(a_0+a_2+a_4+\ldots+a_{2 n}=\frac{3^n+1}{2}\)

Hint

Coefficient of \(x^p\) and \(x^q\) in the expansion of \((1+x)^p\) \({ }^{+q}\) are \({ }^{p+q} \mathrm{C}_p\) and \({ }^{p+q} \mathrm{C}_q\)
and
\(
{ }^{p+q} \mathrm{C}_p={ }^{p+q} \mathrm{C}_q
\)
Hence (a) is the correct answer.

Hint

a is the correct choice. We have
\(
\begin{aligned}
(a+b+c)^n= & {[a+(b+c)]^n } \\
= & a^n+{ }^n \mathrm{C}_1 a^{n-1}(b+c)^1+{ }^n \mathrm{C}_2 a^{n-2}(b+c)^2 \\
& +\ldots+{ }^n \mathrm{C}_n(b+c)^n
\end{aligned}
\)
Further, expanding each term of R.H.S., we note that
First term consist of 1 term.
Second term on simplification gives 2 terms.
Third term on expansion gives 3 terms.
Similarly, fourth term on expansion gives 4 terms and so on.
The total number of terms \(=1+2+3+\ldots+(n+1)\)
\(
=\frac{(n+1)(n+2)}{2}
\)

Hint

(b) is the correct choice. Let \(\mathrm{T}_{r+1}\) be the general term of \(x^2+\frac{2}{x}\), so,
\(
\begin{aligned}
\mathrm{T}_{r+1} & ={ }^{15} \mathrm{C}_r\left(x^2\right)^{15-r} \frac{2}{x} \\
& ={ }^{15} \mathrm{C}_r(2)^r x^{30-3 r} \dots(1)
\end{aligned}
\)
Now, for the coefficient of term containing \(x^{15}\),
\(
30-3 r=15, \quad \text { i.e., } \quad r=5
\)
Therefore, \({ }^{15} \mathrm{C}_5(2)^5\) is the coefficient of \(x^{15}\) (from (1))
To find the term independent of \(x\), put \(30-3 r=0\)
Thus \({ }^{15} \mathrm{C}_{10} 2^{10}\) is the term independent of \(x\) (from (1))
Now the ratio is \(\frac{{ }^{15} \mathrm{C}_5 2^5}{{ }^{15} \mathrm{C}_{10} 2^{10}}=\frac{1}{2^5}=\frac{1}{32}\)

Hint

b is the correct choice. On simplification, we get
\(
z=2{ }^5 \mathrm{C}_0 \frac{\sqrt{3}^2}{2}+{ }^5 \mathrm{C}_2 \frac{\sqrt{3}^3}{2} \frac{i}{2}^2+{ }^5 \mathrm{C}_4 \frac{\sqrt{3}}{2} \frac{i}{2}^4
\)
Since \(i^2=-1\) and \(i^4=1, z\) will not contain any \(i\) and hence \(\mathrm{I}_m(z)=0\).

Hint

\(
\begin{aligned}
& \left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^{15} \text { (Given) }\\
& T_{r+1}={ }^{15} \mathrm{C}_r\left(\frac{3 x^2}{2}\right)^{15-r}\left(-\frac{1}{3 x}\right)^r \text { (standard formula of Tr+1) } \\
& T_{r+1}={ }^{15} \mathrm{C}_r(-1)^r 3^{15-2 r} 2^{r-15} X^{30-3 r} \ldots \dots(i)
\end{aligned}
\)
Now, for \(x\),
\(
\begin{aligned}
& 30-3 r=0 \\
& \rightarrow r=10
\end{aligned}
\)
Substituting the value of \(r\) in eq (i),
\(
\begin{aligned}
T_{r+1} & ={ }^{15} \mathrm{C}_{10} 3^{-5} 2^{-5} \\
& ={ }^{15} \mathrm{C}_{10}\left(\frac{1}{6}\right)^5
\end{aligned}
\)

Hint

\(
\begin{aligned}
\therefore \quad & T_{r+1}={ }^{10} C_r(\sqrt{x})^{10-r}\left(\frac{-k}{x^2}\right)^r \\
& ={ }^{10} C_r(x)^{\frac{1}{2}(10-r)}(-k)^r x^{-2 r} \\
& ={ }^{10} C_r(x)^{5-\frac{r}{2}-2 r}(-k)^r \\
& ={ }^{10} C_r x^{\frac{10-5 r}{2}}(-k)^r
\end{aligned}
\)
For the term free from \(x\),
\(
\frac{10-5 r}{2}=0 \Rightarrow r=2
\)
So, the term free from \(x\) is
\(
\begin{aligned}
& T_{2+1}={ }^{10} C_2(-k)^2 . \\
& \Rightarrow{ }^{10} C_2(-k)^2=405 \\
& \Rightarrow \frac{10 \times 9 \times 8 !}{2 ! \times 8 !}(-k)^2=405 \\
& \Rightarrow 45 k^2=405 \\
& \Rightarrow k^2=9 \quad \therefore k=\pm 3
\end{aligned}
\)

Hint

The given expression is \(\left(1-3 x+7 x^2\right)(1-x)^{16}\).
\(
\begin{aligned}
& =\left(1-3 x+7 x^2\right)\left[{ }^{16} C_0(1)^{16}(-x)^0+{ }^{16} C_1(1)^{15}(-x)+{ }^{16} C_2(1)^{14}(-x)^2+\ldots\right] \\
& =\left(1-3 x+7 x^2\right)\left(1-16 x+120 x^2 \ldots\right)
\end{aligned}
\)
Collecting the term containing \(\mathrm{x}\)
We get \(-16 x-3 x=-19 x\)
Hence, the coefficient of \(x=-19\)

Hint

\(
\begin{aligned}
& \therefore \mathrm{T}_{r+1}={ }^{15} C_r(3 x)^{15-r}\left(\frac{-2}{x^2}\right)^r \\
& ={ }^{15} C_r 3^{15-r} x^{15-3 r}(-2)^r
\end{aligned}
\)
For the term independent of \(x\),
\(
15-3 r=0 \Rightarrow r=5
\)
\(\therefore\) The term independent of \(x\) is
\(
\begin{aligned}
& T_{5+1}={ }^{15} C_5 3^{15-5}(-2)^5 \\
& =\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 !}{5 \times 4 \times 3 \times 2 \times 1 \times 10 !} \cdot 3^{10} \cdot 2^5 \\
& =-3003 \times 3^{10} \times 2^5
\end{aligned}
\)

Hint

Given Expression (i) \((x / a-a / x)^{10}\)
Here index, \(n=10\) (even). So, there is one middle term which is \((10 / 2+1)\) th term, i.e,6th term.
\(
\begin{aligned}
& \therefore T_6=T_{5+1}={ }^{10} C_5\left(\frac{x}{a}\right)^{10-5}\left(\frac{-a}{x}\right)^5 \\
& =-{ }^{10} C_5\left(\frac{x}{a}\right)^5\left(\frac{a}{x}\right)^5 \\
& =-\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 \times 4 \times 3 \times 2 \times 1}\left(\frac{x}{a}\right)^5\left(\frac{x}{a}\right)^{-5} \\
& =-252
\end{aligned}
\)

Hint

Given Expression (ii) \(\left(3 x-x^3 / 6\right)^9\)
Here index, \(n=9\) (odd)
So, there are two middle terms, which are \(((9+1) / 2)\) th i.e, 5 th term and \(((9+1) / 2\) +1)th i.e, 6th term.
\(
\begin{aligned}
& \therefore T_5=T_{4+1}={ }^9 C_4(3 x)^{9-4}\left(-\frac{x^3}{6}\right)^4 \\
& =\frac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 1 \times 5 !} 3^5 x^5 x^{12} 6^{-4} \\
& =\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times \frac{3^5}{3^4 \times 2^4} x^{17} \\
& =\frac{189}{8} x^{17}
\end{aligned}
\)
And \(T_6=T_{5+1}={ }^9 C_5(3 x)^{9-5}\left(-\frac{x^3}{6}\right)^5\)
\(
\begin{aligned}
& =-\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1} \cdot 3^4 \cdot x^4 \cdot x^{15} \cdot 6^{-5} \\
& =-\frac{21}{16} x^{19}
\end{aligned}
\)

Hint

The given expression is \(\left(x-x^2\right)^{10}\)
General Term \(\mathrm{T}_{r+1}={ }^n \mathrm{C}_r^{n-r} y^r\)
\(
\begin{aligned}
& ={ }^{10} C_r(x)^{10-r}\left(-x^2\right) r \\
& ={ }^{10} C_r(x)^{10-r}(-1)^r \cdot\left(x^2\right)^r \\
& =(-1)^r \cdot{ }^{10} C_r(x)^{10-r+2 r} \\
& =(-1)^r \cdot{ }^{10} C_r(x)^{10+r}
\end{aligned}
\)
To find the coefficient of \(x^{15}\)
Put \(10+r=15\)
\(
\begin{aligned}
& \Rightarrow r=5 \\
& \therefore \text { Coefficient of } x^{15}=(-1)^5{ }^{10} C_5 \\
& =-{ }^{10} C_5 \\
& =-252
\end{aligned}
\)
Hence, the required coefficient \(=-252\)

Hint

The given expression is \(\left(x^4-\frac{1}{x^3}\right)^{15}\)
General Term \(T_{r+1}={ }^n C_r x^{n-r} y^r\)
\(
\begin{aligned}
& ={ }^{15} C_r\left(x^4\right)^{15-r}\left(-\frac{1}{x^3}\right)^r \\
& ={ }^{15} C_r(x)^{60-4 r}(-1)^r \cdot \frac{1}{x^{3 r}} \\
& ={ }^{15} C_r(-1)^r \cdot \frac{1}{x^{3 r-60+4 r}} \\
& ={ }^{15} C_r(-1)^r \cdot \frac{1}{x^{7 r-60}}
\end{aligned}
\)
To find the coefficient of \(\frac{1}{x^{17}}\)
Put \(7 r-60=17\)
\(
\begin{aligned}
& \Rightarrow 7 r=60+17 \\
& \Rightarrow 7 r=77 \\
& \therefore r=11
\end{aligned}
\)
Putting the value of \(r\) in the above expression, we get
\(
\begin{aligned}
& ={ }^{15} C_{11}(-1)^{11} \cdot \frac{1}{x^{17}} \\
& =-{ }^{15} C_4 \cdot \frac{1}{x^{17}} \\
& =-\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} \cdot \frac{1}{x^{17}} \\
& =-1365 \cdot \frac{1}{x^{17}}
\end{aligned}
\)
Hence, the coefficient of \(\frac{1}{x^{17}}=-1365\)

Hint

Given expression is \(\left(y^{1 / 2}+x^{1 / 3}\right)^n\)
Binomial coefficient of third term from the end \(=45\)
\(
\begin{aligned}
& \Rightarrow \quad{ }^n C_{n-2}=45 \\
& \Rightarrow \quad{ }^n C_2=45 \\
& \Rightarrow \frac{n(n-1)(n-2) !}{2 !(n-2) !}=45 \\
& \Rightarrow \quad n(n-1)=90 \\
& \Rightarrow \quad n^2-n-90=0 \\
& \Rightarrow(n-10)(n+9)=0 \\
& \Rightarrow \quad n=10[\because n \neq-9]
\end{aligned}
\)
Now, sixth term
\(
\begin{aligned}
& ={ }^{10} C_5\left(y^{1 / 2}\right)^{10-5}\left(x^{1 / 3}\right)^5 \\
& =252 y^{5 / 2} \cdot x^{5 / 3}
\end{aligned}
\)

Hint

General Term \(T(r+1)={ }^n C_r x^{n-r} y^r\)
For coefficient of \((2 r+4)^{\text {th }}\) term, we have
\(
\begin{aligned}
& \mathrm{T}_{2 r+4}=\mathrm{T}_{2 r+3+1} \\
& ={ }^{18} \mathrm{C}_{2 r+3}(1)^{18-2 r-3} \cdot x^{2 r+3} \\
& \therefore \text { Coefficient of }(2 r+4)^{\text {th }} \text { term }={ }^{18} \mathrm{C}_{2 r+3}
\end{aligned}
\)
Similarly, \(\mathrm{T}_{r-2}=\mathrm{T}_{r-3+1}\)
\(
\begin{aligned}
& ={ }^{18} C_{r-3}(1)^{18-r+3} \cdot x^{r-3} \\
& \therefore \text { Coefficient of }(r-2)^{\text {th }} \text { term }={ }^{18} C_{r-3}
\end{aligned}
\)
As per the condition of the questions,
We have \({ }^{18} C_{2 r+3}={ }^{18} C_{r-3}\)
\(
\begin{aligned}
& \Rightarrow 2 r+3+r-3=18 \\
& \Rightarrow 3 r=18 \\
& \Rightarrow r=6
\end{aligned}
\)

Hint

\(
\begin{aligned}
& (1+x)^{2 n}={ }^{2 n} C_0+{ }^{2 n} C_1 x+{ }^{2 n} C_2 x^2+\ldots . .+{ }^{2 n} C_n x^n+\ldots . .+{ }^{2 n} C_{2 n} x^{2 n} \\
& T_2={ }^{2 n} C_1 x, T_3={ }^{2 n} C_2 x^2, T_4={ }^{2 n} C_3 x{ }^3
\end{aligned}
\)
Given, coefficients of \(T_2, T_3\) and \(T_4\) are in A.P.
\(
\begin{aligned}
& \Rightarrow 2 \cdot{ }^{2 n} C_2={ }^{2 n} C_1+{ }^{2 n} C_3 \\
& \Rightarrow \quad 2 \cdot \frac{(2 n) !}{(2 n-2) ! 2 !}=\frac{(2 n) !}{(2 n-1) !}+\frac{(2 n) !}{(2 n-3) ! 3 !} \\
& \Rightarrow(2 n)(2 n-1)=2 n+\frac{2 n(2 n-1)(2 n-2)}{6} \\
& \Rightarrow 6(2 n-1)=6+(2 n-1)(2 n-2) \\
& \Rightarrow 12 n-6=6+4 n^2-6 n+2 \\
& \Rightarrow 4 n^2-18 n+14=0 \\
& \Rightarrow \mathbf2 n^2-9 n+7=0
\end{aligned}
\)

Hint

Given expression is \(\left(1+x+x^2+x^3\right)^{11}\)
\(
\begin{aligned}
& =\left[(1+x)+x^2(1+x)\right]^{11} \\
& =\left[(1+x)\left(1+x^2\right)\right]^{11} \\
& =(1+x)^{11} \cdot\left(1+x^2\right)^{11}
\end{aligned}
\)
Expanding the above expression, we get
\(
\begin{aligned}
& \left({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+{ }^{11} C_3 x^3+{ }^{11} C_4 x^4+\ldots\right) \cdot\left({ }^{11} C_0+{ }^{11} C_1 x^2+{ }^{11} C_2 x^4+\right) \\
& =\left(1+11 x+55 x^2+165 x^3+330 x^4 \ldots\right) \cdot\left(1+11 x^2+55 x^4+\ldots\right)
\end{aligned}
\)
Collecting the terms containing \(x^4\), we get
\(
(55+605+330) x^4=990 x^4
\)
Hence, the coefficient of \(x^4=990\)

Hint

Given expression is \((p / 2+2)^8\)
Since index is \(n=8\) there is only one middle term, i.e., \((8 / 2+1)^{\text {th }}=5^{\text {th }}\) term
\(
\begin{aligned}
& T_5=T_{4+1}={ }^8 C_4\left(\frac{p}{2}\right)^{8-4} \cdot 2^4 \\
& \Rightarrow 1120={ }^8 C_4 p^4 \\
& \Rightarrow 1120=\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 4 \times 3 \times 2 \times 1} p^4 \\
& \Rightarrow \quad 1120=7 \times 2 \times 5 \times p^4 \\
& \Rightarrow \quad p^4=\frac{1120}{70} \\
& \Rightarrow p^4=16 \\
& \Rightarrow p^2=4 \\
& \Rightarrow p=\pm 2
\end{aligned}
\)

Hint

Given, expansion is \(\left(x-\frac{1}{x}\right)^{2 n}\). This Binomial expansion has even power. So, this has one middle term.
i.e., \(\left(\frac{2 n}{2}+1\right)\) th term \(=(n+1) t h\) term
\(
\begin{aligned}
& T_{n+1}={ }^{2 n} C_n(x)^{2 n-n}\left(-\frac{1}{x}\right)^n={ }^{2 n} C_n x^n(-1)^n x^{-n} \\
& ={ }^{2 n} C_n(-1)^n=(-1)^n \frac{(2 n) !}{n ! n !}=\frac{1.2 \cdot 3 \cdot 4 \cdot 5 \ldots .(2 n-1)(2 n)}{n ! n !}(-1)^n \\
& =\frac{1.3 \cdot 5 \ldots(2 n-1) \cdot 2 \cdot 4 \cdot 6 \ldots(2 n)}{12.3 \ldots n(n !)}(-1)^n \\
& =\frac{1.3 .5 \ldots(2 n-1) \cdot 2^n(1.2 \cdot 3 \ldots n)(-1)^n}{(1.2 .3 \ldots n)(n !)} \\
& =\frac{[1.3 .5 \ldots(2 n-1)]}{n !}(-2)^n
\end{aligned}
\)

Hint

In the binomail expansion of \(\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n[(n+1)-7+1]^{\text {th }}\) i.e., \((n-5)^{\text {th }}\) term from the beginning is the \(7^{\text {th }}\) term from the end.
Now,
\(
T_7={ }^n C_6(\sqrt[3]{2})^{n-6}\left(\frac{1}{\sqrt[3]{3}}\right)^6={ }^n C_6 \times 2^{\frac{n}{3}}-2 \times \frac{1}{3^2}
\)
And,
\(
T_{n-5}={ }^n C_{n-6}(\sqrt[3]{2})^6\left(\frac{1}{\sqrt[3]{3}}\right)^{n-6}={ }^n C_6 \times 2^2 \times \frac{1}{3^{\frac{n}{3}}-2}
\)
It is given that,
\(
\begin{aligned}
& \frac{T_7}{T_{n-5}}=\frac{1}{6} \\
& \Rightarrow \frac{{ }^n C_6 \times 2^{\frac{n}{3}}-2 \times \frac{1}{3^2}}{{ }^n C_6 \times 2^2 \times \frac{1}{3^{\frac{n}{3}}-2}}=\frac{1}{6} \\
& \Rightarrow 2^{\frac{n}{3}}-2-2 \times 3^{\frac{n}{3}}-2-2=\frac{1}{6} \\
& \Rightarrow\left(\frac{1}{6}\right)^{4-\frac{n}{3}}=\frac{1}{6} \\
& \Rightarrow 4-\frac{n}{3}=1 \\
& \Rightarrow n=9
\end{aligned}
\)
Hence, the value of \(n\) is 9 .

Hint

We have \((x+a)^n\)
\(
\begin{aligned}
={ }^n C_0 x^n+{ }^n C_1 x^{n-1} a^1+{ }^n C_2 x^{n-2} a^2 \\
+{ }^n C_3 x^{n-3} a^3+\ldots+{ }^n C_n a^n
\end{aligned}
\)
Sum of odd terms,
\(
O={ }^n C_0 x^n+{ }^n C_2 x^{n-2} a^2+\ldots
\)
And sum of even terms,
\(
\begin{gathered}
E={ }^n C_1 x^{n-1} a+{ }^n C_3 x^{n-3} a^3+\ldots \\
\quad \text { Since }(x+a)^n=O+E \cdots \text { (i) } \\
(x-a)^n=O-E \ldots \text { (ii) } \\
\therefore(O+E)(O-E)=(x+a)^n(x-a)^n \\
\Rightarrow O^2-E^2=\left(x^2-a^2\right)^n
\end{gathered}
\)
(ii)
\(
\begin{aligned}
4 O E & =(O+E)^2-(O-E)^2 \\
& =\left[(x+a)^n\right]^2-\left[(x-a)^n\right]^2 \\
& =(x+a)^{2 n}-(x-a)^{2 n}
\end{aligned}
\)

Hint

The general term in the expansion of \(\left(x^2+\frac{1}{x}\right)^{2 n}\) is given by
\(
\begin{aligned}
& T_{r+1}=\cdot{ }^{2 n} C_r \times\left(x^2\right)^{(2 n-r)} \times\left(\frac{1}{x}\right)^r \\
& \Rightarrow T_{r+1}=.{ }^{2 n} C_r \times x^{(4 n-3 r)} . \quad \ldots \text { (i) }
\end{aligned}
\)
This term contains \(x^p\) only when \(4 n-3 r=p\).
And, \(4 n-3 r=p \Rightarrow \frac{(4 n-p)}{3}\).
Putting \(4 n-3 r=p\) in (i), we get
coefficent of \(x^p={ }^{2 n} C_r\), where \(r=\frac{(4 n-p)}{3}\)
\(\frac{(2 n) !}{(r !) \times(2 n-r) !}=\frac{(2 n)}{\left\{\left(\frac{4 n-p}{3}\right) !\right\} \times\left\{\left[2 n-\frac{(4 n-p)}{3}\right] !\right\}}\)
\(\frac{(2 n) !}{\left\{\left(\frac{4 n-p}{3}\right) !\right\} \times\left\{\left(\frac{2 n+p}{3}\right) !\right\}}\)
Hence, the coefficent of \(x^p\) in the expansion of \(\left(x^2+\frac{1}{x}\right)^{2 n}\) is
\(
\frac{(2 n) !}{\left\{\frac{(4 n-p)}{3} !\right\} \times\left\{\left(\frac{2 n+p}{3}\right) !\right\}}
\)

Hint

In the expansion of \(E=\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9\), we have
\(
\begin{aligned}
& T_{r+1}=(-1)^r \cdot{ }^9 C_r \cdot\left(\frac{3}{2} x^2\right)(9-r)\left(\frac{1}{3 x}\right)^r \\
& \Rightarrow T_{r+1}=(-1)^r \cdot{ }^9 C_r \cdot \frac{3(9-2 r)}{2^{(9-r)}} \cdot x^{(18-3 r)} . \\
& \left(1+x+2 x^3\right)\left[\left(a_0 \times \frac{1}{x^3}+a_1 \times \frac{1}{x}+a^2\right) \text { from E }\right] \\
& =\left(1+x+2 x^3\right)\left[\left\{(-1)^7 \cdot{ }^9 C_7 \cdot \frac{3^{-5}}{2^2} \times \frac{1}{x^3}\right\}+\left\{(-1)^6 \cdot{ }^9 C_6 \cdot \frac{3^{-3}}{2^3} \times x^0\right\}\right] \\
& {[\quad x=-1 \Rightarrow 18-3 r=-1 \Rightarrow \quad \mathrm{r} \text { is faction }} \\
& 18-3 r=0 \Rightarrow r=6 \text { and } 18-3 r=-3 \Rightarrow r=7] \\
& =\left(1+x+2 x^3\right)\left[\frac{-1}{27 x^3}+\frac{7}{18}\right] \\
& \therefore \quad \text { required term }\left(\frac{-2}{27}+\frac{7}{18}\right)=\frac{17}{54} .
\end{aligned}
\)

Hint

Given expression is
\(
(x+a)^{100}+(x-a)^{100}
\)
Here, \(n=100\) which is even.
\(\therefore\) Total number of terms \(=\frac{n}{2}+1\) \(=\frac{100}{2}+1=51\)

Hint

Given that \(r>1\) and \(n>2\)
Then \(\mathrm{T}_{3 r}=\mathrm{T}_{3 r-1+1}\)
\(
={ }^{2 n} C_{3 r-1} \cdot x^{3 r-1}
\)
And \(T_{-}(r+2)=T_{r+1+1}\)
\(
={ }^{2 n} C_{r+1} x^{r+1}
\)
We have \({ }^{2 n} C_{3 r-1}={ }^{2 n} C_{r+1}\)
\(
\begin{aligned}
& \Rightarrow 3 \mathrm{r}-1+\mathrm{r}+1=2 \mathrm{n} \quad \ldots \cdot\left[\because{ }^n C_p={ }^n C_q \Rightarrow n=p+q\right] \\
& \Rightarrow 4 \mathrm{r}=2 \mathrm{n} \\
& \mathrm{n}=2 \mathrm{r}
\end{aligned}
\)

Hint

Let the two successive terms in the exansion of \((1+x)^{24}\) be \((r+1)\) th and \((r+2)\) th terms.
Now, \(T_{r+1}={ }^{24} C_r x^r\) and
\(
T_{r+2}={ }^{24} C_{r+1} x^{r+1}
\)
Given that, \(\frac{{ }^{24} C_r}{{ }^{24} C_{r+1}}=\frac{1}{4}\)
\(
\begin{aligned}
& \frac{(24) !}{r !(24-r) !} \\
\Rightarrow & \frac{(24) !}{(r+1) !(24-r-1) !}=\frac{1}{4} \\
\Rightarrow & \frac{(r+1) r !(23-r) !}{r !(24-r)(23-r) !}=\frac{1}{4} \\
\Rightarrow & \frac{r+1}{24-r}=\frac{1}{4} \\
\Rightarrow & 4 r+4=24-r \\
\Rightarrow & r=4 \\
\therefore & T_{4+1}=T_5 \text { and } \\
& T_{4+2}=T_6
\end{aligned}
\)
Hence, \(5^{\text {th }}\) and \(6^{\text {th }}\) terms.

Hint

General Term \(\mathrm{T}_{r+1}={ }^{\mathrm{n}} \mathrm{C}_r x^{n-r} y^r\)
In the expansion of \((1+x)^{2 n}\)
We get \(\mathrm{T}_{r+1}={ }^{2 n} \mathrm{C}_r x^r\)
To get the coefficient of \(x^n\)
Put \(r=n\)
\(\therefore\) Coefficient of \(x^n={ }^{2 n} C_n\)
In the expansion of \((1+x)^{2 n-1}\)
We get \(\mathrm{T}_{r+1}={ }^{2 n-1} C_r x^r\)
\(\therefore\) Coefficient of \(x^n={ }^{2 n-1} C_{n-1}\)
The required ratio is \(\frac{{ }^{2 n} C_n}{{ }^{2 n-1} C_{n-1}}\)
\(
\begin{aligned}
& =\frac{\frac{2 n !}{n !(n !)}}{\frac{(2 n-1) !}{(n-1) !(2 n-1-n+1) !}} \\
& =\frac{\frac{22 n !}{n ! n !}}{\frac{(2 n-1) !}{(n-1)(n !)}} \\
& =\frac{2 n !}{n ! n !} \times \frac{(n-1) ! \cdot n !}{(2 n-1) !} \\
& =\frac{2 n(2 n-1) !}{n ! n(n-1) !} \times \frac{(n-1) ! \cdot n !}{(2 n-1) !} \\
& =\frac{2}{1} \\
& =2: 1
\end{aligned}
\)

Hint

\(
\begin{aligned}
(1+x)^n= & { }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2 \\
& +{ }^n C_3 x^3+\ldots+{ }^n C_n x^n
\end{aligned}
\)
So coefficients of \(2 n d\), 3rd and 4th terms are \({ }^n C_1,{ }^n C_2\), and \({ }^n C_3\) respectively.
Given that, \({ }^n C_1,{ }^n C_2\), and \({ }^n C_3\) are in A.P.
\(
\begin{aligned}
& \therefore 2{ }^n C_2={ }^n C_1+{ }^n C_3 \\
& \Rightarrow 2\left[\frac{n !}{(n-2) ! 2 !}\right]=n+\frac{n !}{3 !(n-3) !} \\
& \Rightarrow 2\left[\frac{n(n-1)}{2 !}\right]=n+\frac{n(n-1)(n-2)}{3 !} \\
& \Rightarrow(n-1)=1+\frac{(n-1)(n-2)}{6} \\
& \Rightarrow \quad 6 n-6=6+n^2-3 n+2 \\
& \Rightarrow n^2-9 n+14=0 \\
& \Rightarrow \quad(n-7)(n-2)=0 \\
& \therefore n=2 \text { or } n=7
\end{aligned}
\)
Since \(n=2\) is not possible, so \(n=7\)

Hint

The coefficient of \(x^n\) in the expansion of \((1+x)^{2 n}\) is \({ }^{2 n} C_n\)
\(
A={ }^{2 n} C_n
\)
The coefficient of \(x n\) in the expansion of \((1+x)^{2 n-1}\) is \({ }^{2 n-1} C_n\)
\(
\begin{aligned}
& \mathrm{B}={ }^{2 n-1} \mathrm{C}_n \\
& \therefore \frac{A}{B}=\frac{{ }^{2 n} C_n}{{ }^{2 n-1} C_n}=\frac{2}{1}=2
\end{aligned}
\)

Hint

Given expression is \(\left(\frac{1}{x}+x \sin x\right)^{10}\)
Number of terms \(=10+1=11\) odd
\(\therefore\) Middle term \(=\frac{11+1}{2}\) th term \(=6^{\text {th }}\) term
\(\mathrm{T}_6=\mathrm{T}_{5+1}\)
\(={ }^{10} C_5\left(\frac{1}{x}\right)^{10-5}(x \sin x)^5\)
\(\therefore{ }^{10} C_5\left(\frac{1}{x}\right)^5 \cdot x^5 \cdot \sin ^5 x=7 \frac{7}{8}\)
\(\Rightarrow{ }^{10} C_5 \cdot \sin ^5 x=\frac{63}{8}\)
\(\Rightarrow \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \cdot \sin ^5 x=\frac{63}{8}\)
\(\Rightarrow 252 \cdot \sin ^5 x=\frac{63}{8}\)
\(\Rightarrow \sin ^5 x=\frac{63}{8 \times 252}\)
\(\Rightarrow \sin ^5 x=\frac{1}{32}\)
\(\Rightarrow \sin ^5 x=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow \sin x=\frac{1}{2}\)
\(
\begin{aligned}
& \Rightarrow \sin \mathrm{x}=\sin \frac{\pi}{6} \\
& \therefore \mathrm{x}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \cdot \frac{\pi}{6}
\end{aligned}
\)

Alternate:

[Hint: \(\mathrm{T}_6={ }^{10} \mathrm{C}_5 \frac{1}{x^5} \cdot x^5 \sin ^5 x=\frac{63}{8} \Rightarrow \sin ^5 x=\frac{1}{2^5} \quad \sin \quad \frac{1}{2}\)
\(
\left.\Rightarrow x=n \pi+(-1)^n \frac{\pi}{6}\right]
\)

Hint

Here \(n=30\) which is even
\(\therefore\) The largest coefficient in \((1+\mathrm{x})^n={ }^n C_{\frac{n}{2}}\)
So, the largest coefficient in \((1+x)^{30}={ }^{30} C_{15}\)
Hence, the value of the filler is \({ }^{30} C_{15}\).

Hint

The expression \((x+y+z)^n\) can be written \(a[x+(y+z)]^n\)
\(
\begin{aligned}
& \therefore[x+y+z]^n={ }^n C_0 x^n(y+z)^0+{ }^n C 1(x) n-1(y+z)+{ }^n C_2(x){ }^{n-2}(y+z)^2+\ldots+{ }^n C n(y+z)^n \\
& \therefore \text { Number of terms } 1+2+3+4+\ldots(n+1) \\
& =\frac{(n+1)(n+2)}{2}
\end{aligned}
\)

Hint

Given \(\left(x^2-1 / x^2\right)^{16}\)
\(
\begin{aligned}
T_{r+1} & ={ }^{16} C_r\left(x^2\right)^{16-r}\left(-\frac{1}{x^2}\right)^r \\
& ={ }^{16} C_r x^{32-4 r}(-1)^r
\end{aligned}
\)
For constant term,
\(
\begin{aligned}
& 32-4 r=0 \Rightarrow r=8 \\
& \therefore T_{8+1}={ }^{16} C_8
\end{aligned}
\)

Hint

The given expansion is \(\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n\)
\(
\begin{aligned}
& \therefore \mathrm{T}_7=\mathrm{T}_{6+1} \\
& ={ }^n \mathrm{C}_6\left(2^{\frac{1}{3}}\right)^{n-6} \cdot \frac{1}{\left(3^{\frac{1}{3}}\right)^6} \\
& ={ }^n \mathrm{C}_6(2)^{\frac{n-6}{3}} \cdot \frac{1}{(3)^2}
\end{aligned}
\)
Now the \(T_7\) from the end \(=T_7\) from the beginning in \(\left(\frac{1}{\sqrt[3]{2}}+\sqrt[3]{2}\right)^n\).
\(
\begin{aligned}
& \therefore \mathrm{T}_7=\mathrm{T}_{6+1} \\
& ={ }^n \mathrm{C}_6\left(\frac{1}{3^{\frac{1}{3}}}\right)^{n-6} \cdot\left(2^{\frac{1}{3}}\right)^6
\end{aligned}
\)
We get \({ }^n C_6(2)^{\frac{n-6}{3}} \cdot\left(\frac{1}{3^2}\right)={ }^n C_6 \frac{1}{\frac{n-6}{3}} \cdot(2)^2\)
\(
\begin{aligned}
& \Rightarrow(2)^{\frac{n-6}{3}} \cdot(3)^{-2}=(3)^{-\left(\frac{n-5}{3}\right)} \cdot(2)^2 \\
& \Rightarrow(2)^{\frac{n-6}{3}} 2 \cdot(3)^{-2+\frac{n-6}{3}}=1 \\
& \Rightarrow 2^{\frac{n-12}{3}} \cdot(3)^{\frac{n-12}{3}}=1 \\
& \Rightarrow(6)^{\frac{n-12}{3}}=(6)^0
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \frac{n-12}{3}=0 \\
& \Rightarrow \mathrm{n}=12
\end{aligned}
\)

Hint

The given expansion is \(\left(\frac{1}{a}-\frac{2 b}{3}\right)^{10}\)
From \(a^{-6} b^4\)
We can take \(r=4\)
\(
\begin{aligned}
& \therefore T_5=T_{4+1} \\
& ={ }^{10} C_4\left(\frac{1}{a}\right)^{10-4}\left(-\frac{2 b}{3}\right)^4 \\
& ={ }^{10} C_4\left(\frac{1}{a}\right)^6\left(\frac{-2}{3}\right)^4 \cdot b^4 \\
& =\frac{10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1} \times \frac{16}{81} \cdot a^{-6} b^4 \\
& =210 \times \frac{16}{81} a^{-6} b^4 \\
& =\frac{1120}{27} a^{-6} b^4
\end{aligned}
\)

Alternate:

\(
\left[\text { Hint : } \quad \mathrm{T}_5={ }^{10} \mathrm{C}_4 \frac{1}{a}^b \frac{-2 b}{3}=\frac{1120}{27} a^{-6} b^4\right]
\)

Hint

Number of term in the expansion \(\left(a^3+b a\right)^{28}\)
\(
\begin{aligned}
& =28+1 \\
& =29 \text { (odd) } \\
& \therefore \text { Middle term }=\frac{29+1}{2}=15^{\text {th }} \text { term } \\
& \therefore T_{15}=T_{14+1} \\
& ={ }^{28} C_{14}\left(a^3\right)^{28-14} \cdot(b a)^{14} \\
& ={ }^{28} C_{14}(a)^{42} \cdot b^{14} \cdot a^{14} \\
& ={ }^{28} C_{14} a^{56} b^{14}
\end{aligned}
\)

Hint

Coefficient of \(x^p\) in the expansion of \((1+x)^{p+q}\) is \({ }^{p+q} C_p\).
Coefficient of \(x^q\) in the expansion of \((1+x)^{p+q}\) is \({ }^{p+q} C_q\)
Now,
\(
\frac{{ }^{p+q} C_p}{p+q C_q}=\frac{\frac{(p+q) !}{p ! q !}}{\frac{(p+q) !}{q ! p !}}=1
\)
Hence, the ratio of the coefficients of \(x^p\) and \(x^q\) in the expansion of \((1+x)^{p+q}\) is \(1: 1\).

Hint

Given: \(\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^2}\right)^{10}\)
Thus, \(T_{r+1}={ }^{10} C_r\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{3}{2 x^2}\right)^r\)
Now, for constant term,
\(10-5 r=0\)
Thus, \(r=2\)
Therefore, the third term is independent of \(x\).

Hint

Given \(25^{15}=(26-1)^{15}\)
\(
\begin{aligned}
& ={ }^{15} C_0 26^{15}-{ }^{15} C_1 26^{14}+\ldots+{ }^{15} C_{14} 26-{ }^{15} C_{15} \\
& =\left({ }^{15} C_0 26^{15}-{ }^{15} C_1 26^{14}+\ldots+{ }^{15} C_{14} 26-13\right)+12
\end{aligned}
\)
Clearly when \(25^{15}\) is divided by 13 , then remainder will be 12 .

Hint

\(
\begin{aligned}
& \sum_{r=0}^{10}{ }^{20} C_r \\
& \left({ }^{20} C_0+{ }^{20} C_1+\ldots+{ }^{20} C_{10}\right) \\
& =\frac{1}{2}\left\{2\left({ }^{20} C_0+{ }^{20} C_1+\ldots+{ }^{20} C_{10}\right)\right\} \\
& =\frac{2}{2}\left\{\left({ }^{20} C_0+{ }^{20} C_1+\ldots+{ }^{20} C_{10}\right)+\left({ }^{20} C_{20}+{ }^{20} C_{19}+\ldots+{ }^{20} C_{10}\right)\right\} \\
& =\frac{1}{2}\left\{\left({ }^{20} C_0+{ }^{20} C_1+\ldots+{ }^{20} C_{10}+{ }^{20} C_{11}+\ldots+{ }^{20} C_{20}\right)+{ }^{20} C_{10}\right\} \\
& =\frac{1}{2}\left\{22^{20}+{ }^{20} C_{10}\right\}=22^{19}+\frac{1}{2}{ }^{20} C_{10}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& 7^9+9^7=(1+8)^7-(1-8)^9 \\
& =\left[{ }^7 C_0+{ }^7 C_1 \cdot 8+{ }^7 C_2(8)^2+{ }^7 C_3(8)^3+\ldots+{ }^7 C_7(8)^7\right]-\left[{ }^9 C_0-{ }^9 C_1 8+{ }^9 C_2(8)^2-{ }^9 C_3(8)^3+\ldots{ }^9 C_9(8)^9\right] \\
& =(7 \times 8+9 \times 8)+\left(21 \times 8^2-36 \times 8^2\right)+\ldots \\
& =(56+72)+(21-36) 8^2+\ldots \\
& =128+64(21-36)+\ldots \\
& =64[2+(21-36)+\ldots]
\end{aligned}
\)
Which is divisible by 64

Hint

Given expression is \(\left[\left(2 x+y^3\right)^4\right]^7=(2 x+3 y)^{28}\)
So, the number of terms \(=28+1=29\)

Hint

\(
(1+x)^{2 n-1}
\)
Here, \(\mathrm{n}\) is an odd number.
Therefore, the middle terms are \(\left(\frac{2 n-1+1}{2}\right)\) th and \(\left(\frac{2 n-1+1}{2}+1\right)\) th, i.e., \(n\)th and \((n+1)\) th terms.
Now, we have
\(
T_n=T_{n-1+1}={ }^{2 n-1} C_{n-1}(x)^{n-1}
\)
And,
\(
T_{n+1}=T_{n+1}={ }^{2 n-1} C_n(x)^n
\)
\(\therefore\) the coefficients of two middle terms are \({ }^{2 n-1} C_{n-1}\) and \({ }^{2 n-1} C_n\).
Now,
\(
{ }^{2 n-1} C_{n-1}+{ }^{2 n-1} C_n={ }^{2 n} C_n
\)
Hence, the sum of the coefficients of two middle terms in the binomial expansion of \((1+x)^{2 n-1}\) is \(^{2 n} C_n\).

Hint

Given that \(3^{400}=(9)^{200}=(10-1)^{200}\)
\(
\begin{aligned}
& \therefore(10-1)^{200}={ }^{200} C_0(10)^{200}-{ }^{200} C_1(10)^{199}+\ldots-{ }^{200} C_{199}(10)^1+{ }^{200} C_{200}(1)^{200} \\
& =10^{200}-200 \times 10^{199}+\ldots-10 \times 200+1
\end{aligned}
\)
So, it is clear that last two digits are 01 .

Hint

Given Binomial expansion is \(\left(x-\frac{1}{x^2}\right)^{2 n}\)
Let \(T_{r+1}\) term is independent of \(x\)
Then, \(T_{r+1}={ }^{2 n} C_r(x)^{2 n-r}\left(-\frac{1}{x^2}\right)^r\) \(={ }^{2 n} C_r x^{2 n-r}(-1)^r x^{-2 r}={ }^{2 n} C_r x^{2 n-3 r}(-1)^r\)
For independent of \(x\)
\(
\begin{aligned}
& 2 n-3 r=0 \\
& \therefore r=\frac{2 n}{3}
\end{aligned}
\)
Which is not a integer
So, the given expansion is not possible

Hint

The given statement is False.
In the given expression \((a+b)^n\), the no. of terms is just 1 more than \(n\), i.e., \(n+1\).