Binomial sum of products of gamma functions

Question

Does anyone know how to prove the following identity for all $k,n \in \mathbb{N}$ \begin{align} &\sum_{i = 0}^k (-1)^i {{k}\choose{i}}\frac{\Gamma(n + i + \frac{1}{2})\Gamma(n + k - i + \frac{1}{2})}{\Gamma(n - i + \frac{1}{2})\Gamma(n - (k - i) + \frac{1}{2})}\\ &= \frac{1}{\pi^2}2^{2k - 1}(1 + (-1)^k)\Gamma\left(\frac{k + 1}{2}\right)^2\cos(n\pi)\Gamma\left(n + \frac{k + 1}{2}\right)\Gamma\left(-n + \frac{k + 1}{2}\right)? \end{align} I am very grateful for any idea or hint!