Closed expression for hypergeometric sum I am trying to simplify an expression and find a closed form for
$$\sum_{m=0}^l \binom{s-m}{s-l} \binom{s-1+m}{s-1}x^m$$
How could I get rid of this summation?
 A: You may argue as GH from MO from your other post.


*

*the coefficient of $y^m$ in $(1-xy)^{-s}$ equals $\binom{s+m-1}{s-1}x^m$;

*the coefficient of $y^{\ell-m}$ in $(1-y)^{\ell-s-1}$ equals $\binom{s-m}{s-\ell}$.


Therefore, the sum on your LHS equals to


*

*the coefficient of $y^{\ell}$ in $(1-xy)^{-s}(1-y)^{\ell-s-1}$.


Unfortunately, this has no closed form. How can we be sure? To this end, denote your sum by
$$f(\ell):=\sum_{m=0}^\ell\binom{s+m-1}{s-1}\binom{s-m}{s-\ell}x^m.$$
As I explained the WZ-method in the other post, the procedure generates a recurrence. However, this time it is a three-term relation
$$(\ell+2)f(\ell+2)+(-sx-\ell x-s+2\ell-x+2)f(\ell+1)+(x-1)(s-\ell)f(\ell)=0$$ 
which reveals that $f(s)$ can not have a closed form.
If you're not interested in the sum, then formulate this as a contour integral. Let $\gamma$ be a closed path (oriented positive) around $z=0$, and apply Cauchy's Integral Formula:
$$f(\ell)=\frac1{2\pi i}\int_{\gamma}\frac{dz}
{z^{\ell+1}(1-xz)^s(1-z)^{s+1-\ell}}.$$
On a positive note, we can derive a generating function for the sequence $f(\ell)$:
$$\sum_{\ell=0}^{\infty}f(\ell)y^{\ell}=\left(\frac{(1+y)^2}{1+y-xy}\right)^s.$$
To see this, start by interchanging summations to proceed as follows:
\begin{align} \sum_{\ell\geq0}f(\ell)y^{\ell}
&=\sum_{m\geq0}\binom{s+m-1}mx^m\sum_{\ell=m}^s\binom{s-m}{\ell-m}y^{\ell} \\
&=\sum_{m\geq0}\binom{s+m-1}mx^my^m(1+y)^{s-m} \\
&=(1+y)^s\sum_{m=0}^{\infty}\binom{s+m-1}m\left(\frac{yx}{1+y}\right)^m \\
&=(1+y)^s\left(1-\frac{yx}{1+y}\right)^{-s} \\
&=\left(\frac{(1+y)^2}{1+y-xy}\right)^s.
\end{align}
A: Mathematica says: 
$$\binom{s}{s-l} \, _2F_1(-l,s;-s;x).$$
(without hypergeometrics for special values of $x,$ like $x=1:$
$$\frac{(-1)^l (l-2 s-1)!}{l! (-2 s-1)!}$$
A: The Mathematica formula mentioned by Rivin is derived via
\begin{equation}
\sum_{m=0}^l \binom{s-m}{s-l}\binom{s-1+m}{s-1}x^m
=
\sum_{m=0}^l \frac{\Gamma(s-m+1)\Gamma(s+m)}{\Gamma(1+s-l)\Gamma(l+1-m)\Gamma(s)}\frac{x^m}{m!}
\end{equation}
\begin{equation}
=
\frac{1}{\Gamma(1+s-l)}\sum_{m=0}^l \frac{\Gamma(s-m+1)(s)_m}{\Gamma(l+1-m)}\frac{x^m}{m!}
\end{equation}
\begin{equation}
=
\frac{1}{\Gamma(1+s-l)}\sum_{m=0}^l (-)^m \Gamma(s+1)\frac{(s)_m}{(-s)_m\Gamma(l+1-m)}\frac{x^m}{m!}
\end{equation}
\begin{equation}
=
\frac{1}{\Gamma(1+s-l)}\sum_{m=0}^l (-)^m \Gamma(s+1)\frac{(s)_m(-l)_m}{(-s)_m(-)^m\Gamma(l+1)}\frac{x^m}{m!}
\end{equation}
\begin{equation}
=
\frac{1}{\Gamma(1+s-l)}\sum_{m=0}^l \Gamma(s+1)\frac{(s)_m(-l)_m}{(-s)_m\Gamma(l+1)}\frac{x^m}{m!}
\end{equation}
\begin{equation}
=
\frac{\Gamma(s+1)}{\Gamma(1+s-l)\Gamma(l+1)}\sum_{m=0}^l \frac{(s)_m(-l)_m}{(-s)_m}\frac{x^m}{m!}
\end{equation}
\begin{equation}
=
\frac{\Gamma(s+1)}{\Gamma(1+s-l)\Gamma(l+1)}{}_2F_1(s,-l;-s;x)
\end{equation}
\begin{equation}
=
\frac{s!}{(s-l)!l!}{}_2F_1(s,-l;-s;x)
\end{equation}
\begin{equation}
=
\binom{s}{l}{}_2F_1(s,-l;-s;x)
\end{equation}
