Question

If $S={\sum }_{0\le }{\sum }_{r<s\le n}(r+s){\left(C_r{-C}_s\right)}^2$, then $S$ equals

• A. $0$
• B. $\left({{}^{2n}C}_n\right)-(n+1)2^{2n}$
• C. $n\left[\right(n+1\left)\right({{}^{2n}C}_n)-2^{2n}]$
• D. $n\left[\right(n+1)2^{2n}-\frac{1}{2}({{}^{2n}C}_n\left)\right]$

Answer 1

Answer: (C)

$n\left[\right(n+1\left)\right({{}^{2n}C}_n)-2^{2n}]$

We have ${(C_0+C_1+C_2+...+C_n)}^2={\sum }_{r=0}{C}_r^2+2P\Rightarrow 2P={\left(2^n\right)}^2-{\sum }_{r=0}{C}_r^2$

But $C_0^2+C_1^2+...+C_n^2{=}^{2n}{C}_n$

Thus, $P=2^{2n-1}-\frac{1}{2}\left({}^{2n}{C}_n\right)$

Next, $Q={(C_0-C_1)}^2+{(C_0-C_2)}^2+...+{(C_0-C_n)}^2$

$+{(C_1-C_2)}^2+{...+(C_1-C_n)}^2+{(C_2-C_3)}^2+...+{(C_2-C_n)}^2+...+{(C_{n-1}-C_n)}^2$

$=n(C_0^2+C_1^2+...+C_n^2)-2P$

But $C_0^2+C_1^2+...+C_n^2{=}^{2n}{C}_n$

and $P=2^{2n-1}-\frac{1}{2}\left({}^{2n}{C}_n\right)$

Next, note that $R$ can also be written as $R=\sum _r^{}\sum _{<s}^{}(r+s)C_r{C}_s=\sum _r^{}\sum _{<s}^{}(n-r+n-s)C_{n-r}{C}_{n-s}$

$=\sum _r^{}\sum _{<s}^{}\left(2n\right)C_r{C}_s-\sum _{r<s}^{}(r+s)C_r{C}_s$

$\Rightarrow 2R=2n\sum _r^{}\sum _{<s}^{}{C}_r{C}_s=2nP$

$\Rightarrow R=\frac{n}{2}\left[2^{2n}{-}^{2n}{C}_n\right]$ $[$See Example 61$]$

Next, $S=\sum _{0\le r<}^{}\sum _{s\le n}^{}(n-r+n-s){(C_{n-r}-C_{n-s})}^2$

$=2n\sum _{0\le r<}^{}\sum _{s\le n}^{}{(C_r+C_s)}^2-S$

$\Rightarrow 2S=2nQ\Rightarrow S=nQ$