SF13

Evaluate $S = \sum_{k=0}^n {n \choose k}^2\,$

Identity: ${n \choose k} = \frac{(-1)^k(-n)_k}{k!}\,$

$S = \sum_{k=0}^n \frac{(-n)_k(-n)_k}{k!k!} = \sum_{k=0}^n \frac{(-n)_k(-n)_k}{(1)_k}\frac{1^k}{k!}\,$

$S = {}_2F_1(-n,-n;1;1)\,$ because the terms in the sum of the generalized hypergeometric function equal 0 when $k>n\,$

Since ${}_2F_1(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\,$

$S = \frac{\Gamma(1)\Gamma(2n+1)}{\Gamma(n+1)\Gamma(n+1)} = \frac{(2n)!}{n!n!} = \frac{2^{2n}\left(\frac{1}{2}\right)_n}{n!}\,$