A second via the Chu-Vandermonde convolution: \begin{equation} \sum_{k=0}^n \binom{x}k \binom{y}{n-k} = \binom{x+y}n \end{equation} \begin{equation} \binom{-1/2}{n} = (-1)^n\binom{2n}{n}\frac 1 {2^{2n}} \end{equation}