<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] 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 $\binom{p}{q}=\frac{p}{p-q}\binom{p-1}{q}=\frac{p(p-1)}{(p-q)(p-q-1)}\binom{p-2}{q}$.
</s>