A Generating function of product of binomial coefficients

Question

Are any of you familiar with the closed form solutions for $$\sum_{j=0}^\infty \binom{j+\alpha+\beta}{j+\alpha} \binom{\beta}{j}x^j $$ where $\alpha$ and $\beta $ are integers?

Thanks!

Answer 1

HINT:

Note that $$(x+2)^n=\sum_{\beta=0}^{n}\sum_{j=0}^{\beta}\binom{n}{n-\beta\space ,\space j\space ,\space \beta-j}x^j$$ in which by $\binom{n}{n-\beta\space ,\space j\space ,\space \beta-j}$ we mean $\frac{n!}{(n-\beta)!\times j!\times (\beta-j)!}$

Now if you let $j+\alpha+\beta=n$ in your question, and notice to this fact that for $j>\beta$, $\binom{\beta}{j}=0$, hence your question simplifies to: $$\sum_{j=0}^{\beta}\binom{n}{n-\beta\space ,\space j\space ,\space \beta-j}x^j$$ which is the inner sum in the first relation.

Answer 2

More generally, consider the Binomial theorem: $$ f(x) = (1+x)^n = \sum_{k=0}^{n} \binom{n}{k}x^k $$ Start taking derivatives (the sum is finite, so you can do it). After $m$ steps (and a bit of algebra) you will get $$ (1+x)^{n-m} \prod_{j={n-m}}^{n} j = m! x^m \sum_{k=m}^{n}\binom{k}{m}\binom{n}{k} x^k $$