Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+\dots +C_n{x}^n,$
then the value of $\underset{0\le r<s\le n}{\sum \sum }{(C_r+C_s)}^2$ is

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

Answer 1

The correct option is D $(n-1)\cdot {}^{2n}{C}_n+2^{2n}$

We have,
$\underset{0\le r<s\le n}{\sum \sum }{(C_r+C_s)}^2$

$={(C_0+C_1)}^2+{(C_0+C_2)}^2+\dots +{(C_0+C_n)}^2+{(C_1+C_2)}^2+{(C_1+C_3)}^2+\dots .+{(C_1+C_n)}^2+⋮+⋮+\cdots +⋮$
$+{(C_{n-2}+C_{n-1})}^2+{(C_{n-2}+C_n)}^2+{(C_{n-1}+C_n)}^2$

$=n(C_0^2+C_1^2+\dots +C_n^2)+2\underset{0\le r<s\le n}{\sum \sum }{C}_r{C}_s$
$=n\cdot {}^{2n}{C}_n+[2^{2n}-{}^{2n}{C}_n]=(n-1)\cdot {}^{2n}{C}_n+2^{2n}$