Problem - 3976
Show that
$$\sum_{i=0}^n \binom{n}{i}^2 = \binom{2n}{n}$$.
This identity can be derived from Vandermonde's identity (# 4274) $$\sum_{i=0}^n \binom{n}{i}^2 =\sum_{i=0}^n \binom{n}{i}\binom{n}{n-i} = \binom{n+n}{n}=\binom{2n}{n}$$
This identity can be derived from Vandermonde's identity (# 4274) $$\sum_{i=0}^n \binom{n}{i}^2 =\sum_{i=0}^n \binom{n}{i}\binom{n}{n-i} = \binom{n+n}{n}=\binom{2n}{n}$$