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?



Closed expression for hypergeometric sum
You may argue as GH from MO from [your other post](https://mathoverflow.net/questions/257349/equality-with-binomials).


* 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](https://mathoverflow.net/questions/257349/equality-with-binomials), 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](https://en.wikipedia.org/wiki/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}


