Question

Value of the expression
$Q={\sum }_{0\le }{\sum }_{r<s\le n}(r+s)C_r{C}_s$ is

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

Answer 1

Answer: (D)

None of these

We have
${(C_0+C_1+....+C_n)}^2={\sum }_{r=0}{C}_r^2+2P$
$\Rightarrow 2P={\left(2^n\right)}^2-{\sum }_{r=0}{C}_r^2$
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
$Q={(C_0-C_1)}^{21}+{(C_0-C_2)}^2+....+{(C_0-C_n)}^2+{\left(C_1{-C}_2\right)}^2+...+{\left(C_1{-C}_n\right)}^2+{\left(C_2{-C}_3\right)}^2+...+{(C_2-C_n)}^2+....+{(C_{n-1}-C_n)}^2$
$=n(C_0^2+C_1^2+....+C_n^2)-2P$
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
$R=\sum {\sum }_{r<s}(r+s)C_r{C}_s$
$=\sum {\sum }_{r<s}(n-r+n-s)C_{n-r}{C}_{n-s}$
$\Rightarrow 2R=2n\sum {\sum }_{r<s}{C}_r{C}_s$
$\Rightarrow R=\frac{n}{2}[2^{2n}-{{}^{2n}C}_n]$
$S={\sum }_{0\le }{\sum }_{r<s\le n}(n-r+n-s){(C_{n-r}-C_{n-s})}^2$
$=2n{\sum }_{0\le }{\sum }_{r<s\le n}{\left(C_r{+C}_s\right)}^2-S$
$\Rightarrow 2S=2nQ\Rightarrow S=nQ$
$T=n{\sum }_{k=0}^n{C_k}^2+2P$
$=(n-1)\left({{}^{2n}C}_n\right)+2^{2n}$