Mathematics - Binomial Theorem Question with Solution | TestHub

$\sum \sum {\left(C_i+C_j\right)}^2=\sum \sum C_i^2+C_j^2+2C_i{C}_j=n\left(C_0^2+C_1^2+\dots \dots +C_n^2\right)+2\sum \sum C_i{C}_j$ is $=n\cdot {}^{2n}{C}_n+2\left(\frac{2^{2n}-{}^{2n}{C}_n}{2}\right)$ Since ${\left(C_0+C_1+\dots +C_n\right)}^2=C_0^2+C_1^2+\dots \dots +C_n^2+2\sum \sum C_i{C}_j$ $2^{2n}={}^{2n}{C}_n+2\sum \sum C_i{C}_j$ ∴ Sum $=n\cdot {}^{2n}{C}_n+2^{2n}-{}^{2n}{C}_n=(n-1{)}^{2n}{C}_n+2^{2n}$