Closed form for $ \sum_{l=0}^{2m} \binom{2m+n-1}{2m-l}x^l $

Question

I am trying to bound an operator and I arrived to a sum of the form $$ \sum_{\ell = 0}^{2m}\binom{2m + n - 1}{2m - \ell}x^{\ell},\qquad \mbox{with}\quad0\le m\le n-1. $$

Does this have a nice closed form ?

EDIT: This (quite obviously) transforms to $$ x^{2m}\sum_{k=0}^{2m} \binom{2m+n-1}{k}x^{-k}. $$ So the question can be rephrased to ask about the closed form of the sum $$ \sum_{k=0}^{m} \binom{m+n}{k}x^{k}. $$

Answer 1

No, it can't have a closed-form for the following reason. It is known that even sum $\sum_{k = 0}^K \binom{N}{k}$ has no closed form. Without loss of generality we can assume that $K \le \frac{N}{2}$. Then

$$\sum_{k = 0}^K \binom{N}{k} = (K \bmod 2)\binom{N}{K} + \sum_{k = 0}^{2\left\lfloor\frac{K}{2}\right\rfloor}\binom{N}{k} = (K \bmod 2)\binom{N}{K} + \sum_{k = 0}^{2m}\binom{2m + n - 1}{k}\\ = (K \bmod 2)\binom{N}{K} + \sum_{\ell = 0}^{2m}\binom{2m + n - 1}{2m - \ell}1^{\ell}$$

for $m = \left\lfloor\frac{K}{2}\right\rfloor$, $n = N - 2m + 1 > \frac{N}{2} \ge 2m \ge m$ and $\ell = 2m - k$. So if your sum had a closed form then $\sum_{k = 0}^K \binom{N}{k}$ would have too. This contradiction shows that your sum has no closed form.

Answer 2

You properly wrote that it depends on the definition of closed form.

If you accept hypergeometric functions to be a "closed" form, you will have $$\sum_{\ell = 0}^{2m}\binom{2m + n - 1}{2m - \ell}x^{\ell}=\binom{2 m+n-1}{2 m} \, _2F_1(1,-2 m;n;-x)$$