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
 A: 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.
A: with the help of Maple i got the following result
$\frac{4^{t-1} (t-x-1) \Gamma \left(t-\frac{1}{2}\right) \Gamma (x+2)
   \binom{t}{x+1}}{\sqrt{\pi } \binom{2 t}{t+x} \Gamma (t+x+1)}$
A: The right-hand side is not correct, but we can show for non-negative integers $0\leq  x\leq t$:
\begin{align*}
\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{1}{2}\left(\frac{x^2+t^2-t}{2t-1}-x\right)\tag{1}
\end{align*}
It is convenient to  use    the coefficient of operator $[z^n]$   to denote the    coefficient of  $z^n$  of  a series.  This  way  we  can  write  for instance
\begin{align*}
\binom{n}{k}=[x^k](1+x)^n\tag{2}
\end{align*}

We start with the left-hand side of (1). Multiplication with $\binom{2t}{t+x}$ gives
  \begin{align*}
\color{blue}{\sum_{n=0}^{\frac{t-x}{2}}}&\color{blue}{n2^{t-2n-x}\binom{t}{n+x}\binom{t-n-x}{t-2n-x}}\\
&=2^{t-x}\sum_{n=1}^{\frac{t-x}{2}}n\binom{t}{n}\binom{t-n}{n+x}\frac{1}{4^n}\tag{3}\\
&=t2^{t-x}\sum_{n=1}^{t}\binom{t-1}{n-1}\binom{t-n}{n+x}\frac{1}{4^n}\tag{4}\\
&=t2^{t-x}\sum_{n=0}^{t-1}\binom{t-1}{n}\binom{t-1-n}{n+x+1}\frac{1}{4^{n+1}}\tag{5}\\
&=t2^{t-x}\sum_{n=0}^{t-1}\binom{t-1}{n}[z^{n+x+1}](1+z)^{t-1-n}\frac{1}{4^{n+1}}\tag{6}\\
&=\frac{t}{4}2^{t-x}[z^{x+1}](1+z)^{t-1}\sum_{n=0}^{t-1}\binom{t-1}{n}\left(\frac{1}{4z(1+z)}\right)^n\tag{7}\\
&=\frac{t}{4}2^{t-x}[z^{x+1}](1+z)^{t-1}\left(1+\frac{1}{4z(1+z)}\right)^{t-1}\tag{8}\\
&=\frac{t}{4^t}2^{t-x}[z^{t+x}](1+2z)^{2t-2}\tag{9}\\
&=\frac{t}{4^t}2^{t-x}\binom{2t-2}{t+x}2^{t+x}\tag{10}\\
&=t\binom{2t-2}{t+x}\\
&=t\binom{2t}{t+x}\frac{(t-x)(t-x-1)}{2t(2t-1)}\tag{11}\\
&=\frac{1}{2}\binom{2t}{t+x}\frac{t^2-2tx+x^2-t+x}{2t-1}\\
&\,\,\color{blue}{=\frac{1}{2}\binom{2t}{t+x}\left(\frac{x^2+t^2-t}{2t-1}-x\right)}
\end{align*}
  and the claim (1) follows.

Comment:


*

*In (3) we use the binomial identity $\binom{n}{t}=\frac{n}{t}\binom{n-1}{t-1}$. We  also set the lower limit to $n=1$ skipping a zero term.

*In (4) we use the binomial identity 
\begin{align*}
\binom{t}{n+x}\binom{t-n-x}{t-2n-x}&=\frac{t!}{(n+x)!(t-n-x)!}\cdot\frac{(t-n-x)!}{(t-2n-x)!n!}\\
&=\frac{t!}{n!(t-n)!}\cdot\frac{(t-n)!}{(n+x)!(t-2n-x)!}\\
&=\binom{t}{n}\binom{t-n}{n+x}
\end{align*}. We also set the upper limit to $n=t$ without changing anything, since we are adding zeros only.

*In (5) we shift the index to start with $n=0$.

*In (6) we apply (2) to the right-hand binomial coefficient.

*In (7) we use  the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (8)  we apply  the binomial  theorem.

*In (9) we do some simplifications and apply the rule as in (7).

*In (10) we select the coefficient of $z^{t+x}$.

*In (11) we apply the binomial identity $\binom{p}{q}=\frac{p}{p-q}\binom{p-1}{q}=\frac{p(p-1)}{(p-q)(p-q-1)}\binom{p-2}{q}$.
