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