Question

If $C_r$ stands for ${}^n{C}_r$ , the sum of the given series $\frac{2\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}{n!}[C_0^2-2C_1^2+3C_2^2-\cdots +(-1{)}^n(n+1)C_n^2]$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

We have $[C_0^2-2C_1^2+3C_2^2-\cdots +(-1{)}^n(n+1)C_n^2]$
$=[C_0^2-C_1^2+C_2^2-\cdots (-1{)}^n{C}_n^2]-[C_1^2-2C_2^2+3C_3^2\cdots -(-1{)}^{n}n{C}_n^2]$
$=(-1{)}^{\frac{n}{2}}\frac{n!}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)}-(-1{)}^{\frac{n}{2}-1}.\frac{1}{2}{n}^n{C}_{\frac{n}{2}}$
$=(-1{)}^{\frac{n}{2}}\frac{n!}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}.(1+\frac{n}{2})$
Therefore the value of the given expression is
$\frac{2\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)}{n!}\times (-1{)}^{\frac{n}{2}}.\frac{n!}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}(1+\frac{n}{2})$
$=(-1{)}^{\frac{n}{2}}(2+n)$