<s>[INST] Proof of an identity involving binomial coefficients I have found numerically that the following identity holds:
\begin{equation}
\sum_{n=0}^{\frac{t-x}{2}} n 2^{t-2n-x}\frac{\binom{t}{n+x}\binom{t-n-x}{t-2n-x}}{\binom{2t}{t+x}} = \frac{x^2+t^2-t}{2t-1},
\end{equation} 
where $n$, $t$, and $x$ are positive integers ($x \leq t$). To make it more visible, values of $n$ range from $0$ to $\frac{t-x}{2}$. 
Any clue about how to prove it?
Thanks, Antonio
 [/INST] This is actually easier then I thought. All what we have to do in here is to use identities of the gamma function along with the Gauss' theorem for the hypergeometric function.
\begin{eqnarray}
&&\sum\limits_{n=0}^{\lfloor \frac{t-x}{2}\rfloor}
n 2^{t-x-2 n} \frac{\binom{t}{n+x} \binom{t-n-x}{t-2 n-x}}{\binom{2 t}{t+x}}=\\
&&\frac{t!(t+x)!}{(2 t)!x!} \sum\limits_{n=0}^{\lfloor \frac{t-x}{2}\rfloor}n 2^{t-x-2 n} \frac{(x-t)^{(2 n)}}{n! (1+x)^{(n)}}=\\
&&\frac{t!(t+x)!}{(2 t)!x!} \cdot \frac{\Gamma(t-x+1)}{\Gamma(\frac{t-x+1}{2})\Gamma(\frac{t-x+2}{2})} \sqrt{\pi} \sum\limits_{n=1}^{\lfloor \frac{t-x}{2}\rfloor} \frac{(\frac{x-t+1}{2})^{(n)} (\frac{x-t}{2})^{(n)}}{(n-1)! (1+x)^{(n)}}=\\
&& \frac{2^{-2+t-x}(x-t)(x-t+1)t!(t+x)!}{\Gamma(1+2 t) \Gamma(2+x)}\cdot F_{2,1}\left[\begin{array}{rr} \frac{x-t+3}{2} & \frac{x-t+2}{2} \\2+x \end{array};1\right]\\
&& \frac{2^{-2+t-x}(x-t)(x-t+1)t!(t+x)!}{\Gamma(1+2 t) \Gamma(2+x)}\cdot \frac{\Gamma(2+x) \Gamma(t-1/2)}{\Gamma(x/2+t/2+1/2)\Gamma(x/2+t/2+1)}=\\
&&\frac{1}{2} \frac{(t-x)(t-x-1)}{2 t-1}
\end{eqnarray}
In the second line from the above we replaced the binomial factors by Pochhammer symbols and simplified the expression. In the third line from the above we used the duplication formula for the Gamma function to reduce the term in sum to Pochhammers symbols with $n$  rather than $2 n$. In the fourth line we used the definition of the hypergeometric function and in the fifth line we used the Gauss' theorem for the hypergeometric function at unity. Finally in the last line we simplified the whole thing again making use of the duplication formula for the gamma function.
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