Question

If $n$ is even, then value of the expression
$C_0-\frac{1}{2}{C}_1^2+\frac{1}{3}{C}_2^2-.....+\frac{{(-1)}^n}{n+1}{C}_n^2$
where
$C_r={{}^{n}C}_r$ is

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

Answer 1

The correct option is D $\frac{{(-1)}^{n/2}n!}{(n+1){\left(n/2\right)!}^2}$

We have
$C_0-\frac{1}{2}{C}_{1}x+\frac{1}{3}{C}_2{x}^2-......+{(-1)}^n\frac{1}{n+1}{C}_n{x}^n\frac{1}{(n+1)x}[1-{(1-x)}^{n+1}]........\left(1\right)$
and
$C_0+C_1\frac{1}{x}+C_2\frac{1}{x^2}+........+C_n{\left(\frac{1}{x}\right)}^n={(1+x)}^n.....\left(2\right)$
Note that
$S=C_0^2-\frac{1}{2}{C}_1^2+\frac{1}{3}{C}_2^2-........+{(-1)}^n\frac{1}{n+1}{C}_n^2$
$=$ Coefficient of constant term in
$[C_0-\frac{1}{2}{C}_{1}x+\frac{1}{3}{C}_2{x}^2-......+{(-1)}^n\frac{1}{n+1}{C}_n{x}^n]\times [C_0+C_1\frac{1}{x}+C_2\frac{1}{x^2}+........+C_n\left(\frac{1}{x^n}\right)]$
$=$ coefficient of constant term in
$=\frac{1}{n+1}\left[\frac{1-{(1-x)}^{n+1}}{x}\right]{(1+\frac{1}{x})}^n$
$=$ coefficient of $c^{n+1}$ in
$\frac{1}{n+1}[{(1+x)}^n-{(1+x^2)}^n(1-x\left)\right]$
$=-\frac{1}{n+1}coefficientof{x}^{n+1}in{(1+x^2)}^n(1-x)$
As $n$ is even, let $n=2m$, then
$S=\frac{1}{2m+1}coefficientof{x}^{2m}in{(1+x^2)}^{2m}$
$=\frac{1}{2m+1}{(-1)}^m\left({{}^{2m}C}_m\right)$
$=\frac{{(-1)}^n}{n+1}\left({{}^{n}C}_{n/2}\right)=\frac{{(-1)}^{n/2}n!}{(n+1){\left(n/2\right)!}^2}$