Question

If $C_r$ stands for ${}^n{C}_r$ then the sum of the series
$\frac{2\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)}{n!}[C_0^2-2C_1^2+3C_2^2-\cdots +(-1{)}^n(n+1)C_n^2]$, where n is an even positive integer is equal to

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

Answer 1

The correct option is B $(-1{)}^{\frac{n}{2}}(n+2)$

Since n is even, let n= 2m then,
L.H.S. = S = $\frac{2\times m!m!}{\left(2m\right)!}[C_0^2-2C_1^2+3C_2^2+\cdots +(-1{)}^{2m}(2m+1)C_{2m}^2]\dots (1)$
or S= $\frac{2\times m!m!}{\left(2m\right)!}\left[\right(2m+1)C_0^2-2m{C}_1^2+(2m-1)C_2^2+\cdots +c_{2m}^2]\dots \left(2\right)$
Adding (1) and (2), we get
$2S=2\frac{m!m!}{\left(2m\right)!}(2m+2)[C_0^2-C_1^2+C_2^2+\cdots +C_{2m}^2]$
We know that $C_0^2-c_1^2-\cdots +C_n^2=(-1{)}^{\frac{n}{2}}{}^n{C}_{\frac{n}{2}}S=2\frac{m!m!}{\left(2m\right)!}(m+1\left)\right[(-1{)}^m{}^{2m}{C}_m]=2(\frac{n}{2}+1)(-1{)}^{\frac{n}{2}}=(-1{)}^{\frac{n}{2}}(n+2)$