Evaluating $\sum_{y=a}^{\infty}{y \choose a} \cdot p^{y-a}$ for $p \in [0,1]$ I am trying to evaluate the following sum $$\sum_{y=a}^{\infty}{y \choose a} \cdot p^{y-a}$$ for $p \in [0,1]$. This looks somewhat like the binomial theorem, but I don't know how I would go about applying it, as the index of summation is $y$ and it's at the top of the binomial coefficient.
I evaluated the sum using Mathematica, and I got $-\frac{(1-p)^{-a}}{p-1}$ which does make it seem like it's been obtained using the binomial theorem, but I am unable to find a way to use it.
Any help would be appreciated.

Let $$c_a=\sum_{y=a}^{+\infty}{y \choose a}p^{y-a}$$
Define the generating function $$\begin{split}
 f(z)&=\sum_{a=0}^{+\infty}c_a z^a\\
&=\sum_{a=0}^{+\infty}\sum_{y=a}^{+\infty}{y \choose a}p^{y-a}z^a\\
&=\sum_{y=0}^{+\infty}\sum_{a=0}^{y}{y \choose a}p^{y-a}z^a\\
&=\sum_{y=0}^{+\infty}p^y\sum_{a=0}^{y}{y \choose a}\left(\frac z p\right)^a\\
&=\sum_{y=0}^{+\infty}p^y\left(1+\frac z p\right)^y\\
&= \frac 1 {1-p-z}\\
&= \frac 1 {1-p} \cdot \frac 1 {1-\frac z { 1-p}}\\
&= \frac 1 {1-p} \sum_{a=0}^{+\infty}\frac 1 {(1-p)^a}z^a
\end{split}$$
This gives the desired formula for $c_a$.
$$c_a=\frac1 {(1-p)^{a+1}}$$
