A way to find a closed form of multinomial convoluted polynome.

It has been a couple of days since I stepped into some hard formula of convoluted polynomial of the form:

$$\sum_k \binom{n}{k}\binom{m}{m-k} x^k$$

I tried to dabble with convolution proofs of Vandermonde to find the nearest resemblance:

$$\sum_k \sum_{l=0}^k \binom{n}{l}\binom{m}{k-l} x^k = \sum_k \binom{n+m}{k}x^k$$

Also I did my research in the book Combinatorial identities of John Riordan I have'nt found any proof to any closed expression of this form.

Can any one give me a link or helping book or any primary clue/hint toward a solution ? Thank you very much.

Solution 1:

Let $(s)_n=\Gamma(s+n)/\Gamma(s)$. For $n\in\Bbb Z$, $(s)_n$ admits the transformations $$ (s)_n=(-1)^n(1-s-n)_n=\frac{(-1)^n}{(1-s)_{-n}}. $$ From this it's easy to show $$ \binom{n}{k}=\frac{n!}{k!(n-k)!}=(-1)^k\frac{(-n)_k}{k!}=(-1)^k\frac{(-n)_k}{(1)_k}. $$ Hence, the sum in question becomes $$ \begin{align} S &=\sum_{k=0}^\infty\binom{n}{k}\binom{m}{m-k}x^k\\ &=\sum_{k=0}^\infty(-1)^k\frac{(-n)_k}{(1)_k}\binom{m}{m-k}x^k. \end{align} $$ Likewise, $$ \binom{m}{m-k}=\frac{m!}{(m-k)!k!}=\binom{m}{k}=(-1)^k\frac{(-m)_k}{k!}. $$ And so, $$ \begin{align} S &=\sum_{k=0}^\infty \frac{(-n)_k(-m)_k}{(1)_k}\frac{x^k}{k!}={_2F_1}\left({-n,-m\atop 1};x\right). \end{align} $$ Note that if $n\in\Bbb N_0$ or $m\in\Bbb N_0$ then this hypergeometric function is degenerate and truncates to a polynomial in $x$.