Question

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

• A. $\frac{1}{2}[2^{2n}-{}^{2n}{C}_n]$
• B. $\frac{1}{4}[2^{2n}-{}^{2n}{C}_n]$
• C. $\frac{1}{2}[2^{2n}+{}^{2n}{C}_n]$
• D. $\frac{1}{2}[2^n-{}^{2n}{C}_n]$

Answer 1

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

We have,

$\sum _{r=0}^n\sum _{s=0}^n{C}_r{C}_s=\left(\sum _{r=0}^n{C}_r^2\right)+2\underset{0\le r<s\le n}{\sum \sum }{C}_r{C}_s$

$\Rightarrow 2^{2n}={}^{2n}{C}_n+2\underset{0\le r<s\le n}{\sum \sum }{C}_r{C}_s$

$\Rightarrow \underset{0\le r<s\le n}{\sum \sum }{C}_r{C}_s=\frac{1}{2}[2^{2n}-{}^{2n}{C}_n]$