Question

Value of $\sum {\sum }_{0\le r<s\le n}{(r+s)(C_r+C_s)}^2$ is

• A. $n\left[\right(n-1\left)\right({{}^{2n}C}_n)+2^{2n}]$
• B. $n\left[\right(n+1\left)\right({{}^{2n}C}_n)+2^{2n}]$
• C. $n[2^{2n}-n({{}^{2n}C}_n\left)\right]$
• D. None of these

Answer 1

The correct option is A $n\left[\right(n-1\left)\right({{}^{2n}C}_n)+2^{2n}]$

$S={\sum }_{0\le r}{\sum }_{<s\le n}{(n-r+n-s)(C_{n-r}-C_{n-s})}^2$
$=2n{\sum }_{0\le r}{\sum }_{<s\le n}{(C_r+C_s)}^2-S$
$\Rightarrow S=n\left[\right(n-1\left)\right({{}^{2n}C}_n)+2^{2n}]$