Question

If $C_r$ stands for ${}^n{C}_r$, then the sum of the series $\frac{2\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}{n!}[C_0^2-2C_1^2+3C_2^2-....+(-1{)}^n(n+1)C_n^2]$, 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,
$C_0^2-2C_1^2+3C_2^2+4C_3^2+.....+(-1{)}^n(n+1)C_n^2$
$=[C_0^2-C_1^2+C_2^2-C_3^2+...+(-1){}^n{C}_n^2]$
$-[C_1^2-2C_3^2+3C_3^2-....+(-1){}^n{C}_n^2]$
$=(-1{)}^{n/2}\frac{n!}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)}-(-1{)}^{\frac{n}{2}-1}\frac{n}{2}\frac{n!}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}$
$=(-1{)}^{n/2}\frac{n!}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}(1+\frac{n}{2})$
$\therefore \frac{2\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}{n!}[C_0^2-2C_1^2+3C_2^2-.....+(1{)}^r(n+1)C_n^2]$
$=\frac{2\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}{n!}(-1{)}^{n/2}\frac{n!}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}\frac{(n+2)}{2}=(-1{)}^{n/2}(n+2)$