Can somebody help me in proving the following equation?
\begin{align*}&\textstyle \sum _{d=0} ^{n} \frac{1}{d!(n-d)!} \frac{\Gamma (b+d) \Gamma (b+n-d) \Gamma (c-n+d) \Gamma (c-b+1-n + 2d) \Gamma (-c-d) \Gamma (-c-b+1 +n -2d)}{\Gamma (b) \Gamma (b) \Gamma (c-n+2d) \Gamma (c-b+1-n+d) \Gamma (-c+n-2d) \Gamma (-c-b+1-d)} \notag \\ &= \frac{1}{n!} \frac{\Gamma(2b+n)}{\Gamma(2b)}, \end{align*} where $n$ is an integer, $b$ and $c$ are complex numbers. In particular, the dependence on $c$ vanishes.
I explicitly checked this up to $n=7$ order by mathematica, so I believe this equation indeed holds. But I do not know how to prove this for now. Are there special identities that allow this equation to hold?
If I manipulate the equation a little bit, I get the following equation.
\begin{align*} 1 = \textstyle \sum _{d=0} ^{n} \frac{n!}{d!(n-d)!} \frac{B (b+d, b+n-d) B (c-n+d, -c-b+1+n-2d) B (c-b-n+2d+1, -c-d)}{B (b,b) B (c-n+2d,-c-b-d+1) B (c-b-n+d+1, -c+n-2d)} \end{align*}
Now both $b$ and $c$ dependencies vanish.
The proof by induction will not be useful because I want to do a similar manipulations to more complicated combinations of gamma functions.
Thanks!