Question

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

• A. $n[2^{2n-1}-{}^{2n-1}{C}_{n-1}]$
• B. $n[2^{2n-1}+{}^{2n-1}{C}_{n-1}]$
• C. $2n[2^{2n-1}-{}^{2n-1}{C}_{n-1}]$
• D. $2n[2^{2n-1}+{}^{2n-1}{C}_{n-1}]$

Answer 1

The correct option is A $n[2^{2n-1}-{}^{2n-1}{C}_{n-1}]$

We have,
$\sum _{r=0}^n\sum _{s=0}^n(r+s)C_r{C}_s$
$=\sum _{r=0}^{n}2r(C_r{)}^2+2\underset{0\le r<s\le n}{\sum \sum }(r+s)C_r{C}_s$
$=2\sum _{r=0}^{n}r(C_r{)}^2+2\underset{0\le r<s\le n}{\sum \sum }(r+s)C_r{C}_s$
$\Rightarrow n\cdot 2^{2n}=2(\frac{n}{2}\cdot {}^{2n}{C}_n)+2\underset{0\le r<s\le n}{\sum \sum }(r+s)C_r{C}_s$
$\Rightarrow \underset{0\le r<s\le n}{\sum \sum }(r+s)C_r{C}_s=\frac{1}{2}[n\cdot 2^{2n}-n\cdot {}^{2n}{C}_n]$
$=\frac{n}{2}[2^{2n}-\frac{2n}{n}\cdot {}^{2n-1}{C}_{n-1}]$
$=n[2^{2n-1}-{}^{2n-1}{C}_{n-1}]$