How can I sum the series $e^{-2}\frac{(3)^n}{n!}\sum_{k=0}^{\infty}\left ( \frac{1}{2}\right )^k\frac{1}{(k-n)!}$

Question

How can I sum the following series?

$$e^{-2}\frac{(3)^n}{n!}\sum_{k=0}^{\infty}\left ( \frac{1}{2}\right )^k\frac{1}{(k-n)!}$$

I think I can make this sum in the form of exponential expansion but not able to think how. Any initial hint would be great. Thanks in advance.

Answer 1

Basically

$$\sum_{n=k}^{\infty}\frac{x^n}{(n-k)!}=x^k\sum_{n=k}^{\infty}\frac{x^{n-k}}{(n-k)!}=x^k\sum_{n=0}^{\infty}\frac{x^{n}}{n!} = x^ke^x$$

now take $x=\frac12$