Question

If ${(1+x)}^n=C_0+C_{1}x+...+C_n{x}^n$, then the value of $\sum _{0\le r<}^{}\sum _{s\le n}^{}(C_r+C_s)$ is equal to

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

Answer 1

Answer: (C)

$(n-1)2^n$

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

$\Rightarrow (n+1)2^{n+1}=2\left(\sum _{r=0}^n{C}_r\right)+2\sum _{0\le r<}^{}\sum _{s\le n}^{}(C_r+C_s)$

$\Rightarrow (n+1)2^{n+1}=2.2^n+2\sum _{0\le r<}^{}\sum _{s\le n}^{}(C_r+C_s)$

$\Rightarrow \sum _{0\le r<}^{}\sum _{s\le n}^{}(C_r+C_s)=(n-1)2^n$