Question

If for $n\in N$, ${\sum }_{k=0}^{2n}{(-1)}^k{\left({{}^{2n}C}_k\right)}^2=A$, then value of ${\sum }_{k=0}^{2n}{(-1)}^k(k-2n){\left({{}^{2n}C}_k\right)}^2$ is

• A. $nA$
• B. $-nA$
• C. $0$
• D. none of these

Answer 1

The correct option is B $-nA$

Let
${\sum }_{k=0}^{2n}{(-1)}^k(k-2n){\left({{}^{2n}C}_k\right)}^2.........\left(1\right)$
$\Rightarrow S=-{\sum }_{k=0}^{2n}{(-1)2n-}^k(2n-k){\left({{}^{2n}C}_{2n-k}\right)}^2$
Writing the terms in $S$ in the reverse order, we get
$S=-{\sum }_{k=0}^{2n}{(-1)}^{k}k{\left({{}^{2n}C}_k\right)}^2.....\left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$ we get
$2S=-2n{\sum }_{k=0}^{2n}{(-1)}^k{\left({{}^{2n}C}_k\right)}^2=-2nA$
$\Rightarrow S=-nA$