Question

If $\frac{C_0}{1}-\frac{C_1}{3}+\frac{C_2}{5}-....+\frac{{{(-1)}^{n}C}_r}{2n+1}=k{\int }_0^1{x(1-x^2)}^{n-1}dx$ then $k=$ is

• A. $\frac{2^{n+1}.n.{(n!)}^2}{(n+1)!}$
• B. $\frac{2^{n-1}.n.{(n!)}^2}{(n-1)!}$
• C. $\frac{2^{2n-1}.n.{(n!)}^2}{(2n-1)!}$
• D. $\frac{2^{2n+1}.n.{(n!)}^2}{(2n+1)!}$

Answer 1

The correct option is D $\frac{2^{2n+1}.n.{(n!)}^2}{(2n+1)!}$

$\because \frac{C_0}{1}-\frac{C_1}{3}+\frac{C_2}{5}-\frac{{(-1)}^n{C}_n}{2n+1}=k{\int }_0^{1}x{(1-x^2)}^{n-1}dx$
$\Rightarrow {\int }_0^1{(1-x^2)}^{n}dx=k{\left[\frac{{(1-x^2)}^n}{-2n}\right]}_0^1=\frac{k}{2n}$
Put $x=\sin \theta ,dx=\cos \theta d\theta$
$\therefore {\int }_0^{\frac{\pi }{2}}{\cos }^{2n+1}\theta d\theta =\frac{k}{2n}$
$\Rightarrow \frac{2n(2n-2)(2n-4)....4.2}{(2n+1)(2n-1)(2n-3)....5.3.1}=\frac{k}{2n}$
$\Rightarrow \frac{{[2n(2n-2)(2n-4).....4.2]}^2}{(2n+1)2n(2n-1)(2n-2).....5.4.3.2.1}=\frac{k}{2n}$
$\Rightarrow \frac{2^{2n}{(n!)}^2}{(2n+1)!}=\frac{k}{2n}\Rightarrow k=\frac{2^{2n+1}n{(n!)}^2}{(2n+1)!}$