Question

The value of $\sum _{0\le i<}^{}\sum _{j\le n}^{}({}^n{C}_i+{}^n{C}_j)is:$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$\sum _{0\le i<}^{}\sum _{j\le n}^{}({}^n{C}_i+{}^n{C}_j)$

$\{\begin{array}{c}{C}_0+C_1+C_2+...+C_n\\ +C_1+C_2+...+C_n\\ +C_2+C_3+...+C_n\\ .........\\ +C_{n-1}+C_n.\end{array}$

$=(C_0+2C_1+3C_2+...+{}^n{C}_{n-1}+{}^{n+1}{C}_n-C_n.....(1)$

$C_{0}x+C_1{x}^2+C_2{x}^3+....+{}^n{C}_{n-1}+C_n{x}^{n+1}=x(1+x{)}^n$

Diff. both sides , we get

$C_0+2C_1+3C_{+}...+(n+1)C_n{x}^n$

$=(1+x{)}^n+xn(1+x{)}^{n-1}$

Putting x = 1 ,

$C_0+2C_1+3C_2+....+(n+1)C_n=2^n+n{2}^{n-1}$

= $(n+2)2^{n-1}$

Hence , $\sum _{0\le i<}^{}\sum _{j\le n}^{}(C_i+C_j)$ = $(n+2)2^{n-1}-1$