Vandermonde's identity and the close form of $\sum_{k=0}^r C(n,k) C(m,r-k) x^k$

Question

I have a question related to Vandermonde's identity:

From Vandermonde's identity, we have: $$ \binom{n+m}{r}=\sum_{k=0}^r \binom{n}{k}\binom{m}{r-k} $$ Now, I have an extra term $x^k$ inside the sum, i.e. $$ f(x)=\sum_{k=0}^r \binom{n}{k}\binom{m}{r-k}x^k $$

Is there any closed form solution available for $f(x)$? I would be grateful for any hint?

Thanks.

Answer 1

Hint: A closed form is presumably not available. But, we can write it as the coefficient of a polynomial, which might be useful. Note, that $f(x)$ is dependent on $r$ and should be parametrized with it.

The following holds

\begin{align*} (1+xy)^n(1+y)^m&=\sum_{k=0}^n\binom{n}{k}(xy)^k\sum_{l=0}^m\binom{m}{l}y^l\\ &=\sum_{r=0}^{\infty}\left(\sum_{{k+l=r}\atop{k,l\geq 0}}\binom{n}{k}\binom{m}{l}(xy)^ky^l\right)\\ &=\sum_{r=0}^{\infty}\left(\sum_{k=0}^{r}\binom{n}{k}\binom{m}{r-l}x^k\right)y^r\tag{1}\\ &=\sum_{r=0}^{\infty}f(x;r)y^r \end{align*}

If we denote with $[x^r]$ the coefficient of $x^r$ in a series, we can write according to (1) for $r\geq 0$

\begin{align*} f(x;r)=[y^r](1+xy)^n(1+y)^m\qquad\qquad m,n\geq 0 \end{align*}