Question: I am trying to evaluate a sum consisting of a possion distribution, multiplied by a binomial distribution. I already know by the power of maple that 

> 
> $$ \sum\_{n=k}^\infty \frac{(\lambda\cdot s)^n}{n!}e^{-\lambda s}
>  \binom{n}{n-k}p^{n-k}(1-p)^k = \frac{(s\lambda)^ke^{-s\lambda}}{k!} (1-p)^k e^{s\lambda p} $$
> 
> 
> 

Where the answer also could be written as $P\_r(s\lambda)(1-p)^k e^{s\lambda p}$.
My guess is that I have to split the sum into $[k,\infty)=[0,\infty)-[0,k-1]$. 
Also the sum reminds me of one of the definitions of $e$, and the binomial formula
$$
e^x = \sum\_{k}\frac{x^k}{k!} \ , \qquad (x+y)^n=\sum\_k \binom{n}{k}x^{n-k}y^k 
$$
Where both sums goes from $0$ to $\infty$. However I am not quite able to see how to use this to evalutate the sum. Any help or advice is greatly appreaciated. 

For the curious I am trying to solve **6b** from [here](http://www.math.ntnu.no/~karikriz/TMA4240-H13/ov6-e.pdf), it should be 
$$P(Y=k) = \sum\_{n=k}^\infty P(Y=k \cap N=n) = \frac{(s\lambda)^ke^{-s\lambda}}{k!} (1-p)^k e^{s\lambda p}$$

Answer: 
Factoring the term $\displaystyle (\lambda s)^k\mathrm e^{-\lambda s}\frac1{k!}(1-p)^k$ which is independent of $n$, one is left with the series
$$
(\lambda s)^k\mathrm e^{-\lambda s}\frac1{k!}(1-p)^k\cdot\sum\_{n\geqslant k}(\lambda s)^{n-k}\frac1{(n-k)!}p^{n-k}.
$$
You might want to finish, identifying the last sum...
Notice that it can be written in the form : $$ \frac{(1-p)^k e^{- \lambda s} p^{-k}}{k!} \sum\_{n=k}^{\infty} \frac{(\lambda s p)^n}{(n-k)!} $$
Do a simple manipulation to $$ \sum\_{n=k}^{\infty} \frac{(\lambda s p)^n}{(n-k)!} $$ and then use the identity : $$ \sum\_{n=0}^{\infty} \frac{x^n}{n!} = e^x $$