I've been trying to find the MGF for the following formula:
$$ {{r+x-1}\choose {x}} p^r(1-p)^x$$
with the following constraints $r>0$ and $0<p<1$ and finally x is discrete $x = 0,1,2,3...$
the MGF is:
$$ E(e^tx) = \sum_{x=0}^{\infty}e^{tx}{{r+x-1}\choose {x}} p^r(1-p)^x $$
and then I get the following:
$$ p^r\sum_{x=0}^{\infty}{{r+x-1}\choose {x}} ((1-p)e^t)^x$$
I'm kinda stuck here, how do I evaluate this sum? In specific, how do I evaluate the combination part, as $((1-p)e^t)^x = \frac{p}{1-(1-p)e^t}$ but the sum of the combination remains and I'm not too sure how to proceed. Any advice would be appreciated.
Thank you
Hint:
$(-1)^x \binom{r+x-1}{x} = \binom{-r}{x}$