Question

If $C_r$ stands for $ {}^nC_r$, then the sum of the series $\frac{2\bigg(\frac{n}{2}\bigg)!\bigg(\frac{n}{2}\bigg)!}{n!} [C_0^2-2C_1^2+3C_2^2-....+(-1)^n(n+1)C_n^2]$, where $n$ is an even positive integer, is

• A. $(-1)^{n/2}(n+2)$
• B. $(-1)^n(n+1)$
• C. $(-1)^{n/2}(n+1)$
• D. None of these

Answer 1

Answer: (A)

We have,
$C_0^2-2C_1^2+3C_2^2+4C_3^2+.....+(-1)^n(n+1)C_n^2$
$ =[C_0^2-C_1^2+C_2^2-C_3^2+...+(-1) \,{}^nC_n^2]$
$ -[C_1^2-2C_3^2+3C_3^2-....+(-1) \,{}^nC_n^2]$
$ =(-1)^{n/2} \frac{n!}{\bigg(\frac{n}{2}\bigg)!\bigg(\frac{n}{2}\bigg)} -(-1)^{\frac{n}{2}-1} \frac{n}{2}\frac{n!}{\bigg(\frac{n}{2}\bigg)!{\bigg(\frac{n}{2}\bigg)!}}$
$=(-1)^{n/2} \frac{n!}{\bigg(\frac{n}{2}\bigg)!{\bigg(\frac{n}{2}\bigg)!}} \bigg(1+\frac{n}{2}\bigg)$
$\therefore \, \frac{2 \bigg(\frac{n}{2}\bigg)!\bigg(\frac{n}{2}\bigg)!}{n!} [C_0^2-2C_1^2+3C_2^2 -.....+(1)^r(n + 1)C^2_n] $
$=\frac{2\bigg(\frac{n}{2}\bigg)!\bigg(\frac{n}{2}\bigg)!}{n!}(-1)^{n/2} \frac{n!}{\bigg(\frac{n}{2}\bigg)! \bigg(\frac{n}{2}\bigg)!}\frac{(n+2)}{2}=(-1)^{n/2}(n+2)$