$\sum_{x=0}^{\infty}\frac{(r+x+m-1)!}{(r+m-1)!x!}(1-p)^{x}$

Question

How to simplify given series

$$\sum_{x=0}^{\infty}\frac{(r+x+m-1)!}{(r+m-1)!x!}(1-p)^{x}$$

My solution:

$y=r+m-1$

$q=1-p$

$$\sum_{x=0}^{\infty}\frac{(y+x)!}{y!x!}(1-p)^{x}$$

$1+(y+1)q+\frac{(y+2)(y+1)q^2}{2}+......$

$(1-q)^{-(y+1)}$

Answer 1

Hint: Consider the expansion of $(1-t)^{-n}$, using the Newton generalized binomial theorem, or equivalently the Maclaurin series.