Question

If ${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+...+C_n{x}^n$, then $\sum _{0\le i\le }^{}\sum _{j\le n}^{}{(C_i+C_j)}^2=$

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

Answer 1

Answer: (A)

$(n-1){.}^{2n}{C}_n+2^{2n}$

${\sum }_{0\le i\le }{\sum }_{j\le n}{(C_i+C_j)}^{2}i=0,1,2,...,(n-1);j=1,2,3,...,n$ and $i<j$

$=n(C_0^2+C_1^2+...+C_n^2)+2\sum \sum C_i{C}_{j}0\le i\le j\le n=n{.}^{2n}{C}_n+[{(C_0+C_1+...+C_n)}^2-(C_0^2+C_1^2+...+C_n)]=n{.}^{2n}{C}_n+{\left(2^n\right)}^2{-}^{2n}{C}_n=(n-1){.}^{2n}{C}_n+2^{2n}$