TITLE: An identity of complicated combinations of gamma functions (related to hypergeometric functions)
QUESTION [6 upvotes]: 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!

REPLY [2 votes]: (too long or too complicated as a comment). 
There are at least two relevant properties in your problem.
The part of the expression inside your initial sum that does not depend on $c$ is already equal to the result.
\begin{align*}\sum _{d=0} ^{n}    \frac{1}{d!(n-d)!} \frac{\Gamma (b+d) \Gamma (b+n-d)}{\Gamma (b) \Gamma (b)} 
= \frac{1}{n!} \frac{\Gamma(2b+n)}{\Gamma(2b)} = \frac{1}{n!}\prod_{i=0}^{n-1}(2b+i),
\end{align*}
a binomial sum (or a finite product if you like) you can manipulate in many ways.
You succeeded in finding a one-arbitrary-complex-parameter weight expression that leaves that sum invariant. It is a very interesting property that can be exploited if one sees this sum as a discrete probability distribution. The sum without the $c$ fractions is probably a limit when $c$ approaches an integer.
Another related point is that the original summand is invariant by $d\rightarrow n-d$ so you would always obtain a symmetric distribution.