I try to find a closed form of the following sum of binomials:
$$\sum_{l=0}^k(-1)^l \binom{m}{k-l}\binom{n+k-1}{l},$$
where $k$, $m$, $n$ are all non-negative integers but do not have any other relation.
Is there any identity that can be useful here?
If we start with a Chebyshev polynomial of the second kind $$ x^{n+k}\cdot U_{n+k}\left(\frac{x}{2}\right) = x^{2n}\sum_{2l\leq (n+k)}(-1)^l\binom{n+k-l}{l}(x^2)^{k-l}\tag{1}$$ we get that the initial sum is the coefficient of $x^{2k}$ in the product $$ (1+x^2)^m\cdot x^{k-n}\cdot U_{n+k}\left(\frac{x}{2}\right)\tag{2}$$ or the coefficient of $x^{n+k}$ in $(1+x^2)^m\cdot U_{n+k}\left(\frac{x}{2}\right)$. I fear this does not simplify much further.