Let $a>0, b>0,m>0$

$H(t)=\sum\limits_{k=0}^{\infty} {k \choose a}{m-k \choose b}t^k$

So what is the closed form of $H(t)$?

What I know currently is:

$\sum\limits_{0 \le k \le t} {t-k \choose r} {k \choose s} = {t+1 \choose r+s+1}$

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What makes you think that there is a closed form? What do you know about hypergeometric functions? –  Phira Oct 12 '11 at 15:36
I don't seem to be able to convince the site to interpret correctly a link into W|A... but do enter the text «sum[Binomial[k,a] Binomial[m-k,b] t^k, {k, 0, infinity}]» into Wolfram Alpha. –  Mariano Suárez-Alvarez Oct 12 '11 at 16:03
As always, you could turn all the binomial coefficients into Pochhammer symbols to obtain the hypergeometric representation... –  Guess who it is. Oct 12 '11 at 17:22
Fan Zhang, as I observed above your series sums to a well-known hypergeometric function. Unless you add extra conditions on $a$, $b$ and $m$, there is really nothing more to be said, and anyone will tell you that the expression $\binom{0}{a} \binom{m}{b} \, _3F_2(1,1,b-m;1-a,-m;t)$ which Wolfram Alpha returns is the closed form. –  Mariano Suárez-Alvarez Oct 15 '11 at 5:03
FanZhang, I suggest you read and ponder this page. It might help you get the point (made by others in the comments) that very few functions can be written with the functions you know, and that $H$ is not one of them (for general values of the parameters $a$, $b$ and $m$). –  Did Oct 15 '11 at 5:54

$$H(t)=-\frac1{4\pi^2}\oint_{|z|=1}\oint_{|w|=1}\frac{(1+w)^{m+1}-(1+z)^{m+1}t^{m+1}}{1+w-(1+z)t}\,\frac{\mathrm dw}{w^{b+1}}\,\frac{\mathrm dz}{z^{a+1}}$$