# Computing $\sum_{m=0}^{l} {{k-l}\choose{m}}{l\choose m}x^m$

Given that $k \ge 0$ and $0 \le m \le l \le k$, how does one simplify the expression \begin{align*} \sum_{l=0}^k (-1)^l\left[\sum_{m=0}^{l} {{k-l}\choose{m}}{l\choose m}x^m\right] u^{k-l} v^l, \end{align*} given that $x$ here is actually a variable that depends on $u$ and $v$?

The hard part for me is simplifying the inner sum, which looks like Vandermonde identity but it is not the same. I have tried to consider generating function but I am stuck as well. Any idea would be highly appreciated.

• Actually if you want $(u-v-x)^n= \sum^n_{i,j,k}(^n_{i,j,k})u^iv^jx^k$ then you need something close to the summation that you have. – Sentinel135 Mar 10 '17 at 7:20
• Have you tried generating functions with a new variable $y$? – tehjh Mar 10 '17 at 7:44

$$\sum_{m=0}^{l} {{k-l}\choose{m}}{l\choose m}x^m=P_l^{(0,-k-1)}(1-2 x)$$ where appears the Jacobi polynomial.
It could also be expressed as $$\sum_{m=0}^{l} {{k-l}\choose{m}}{l\choose m}x^m=\, _2F_1(-l,l-k;1;x)$$ where appears the Gaussian or ordinary hypergeometric function.