Finding $\mathbb{E}(X+1)$ and $\mathrm{var}(X+1)$ of a Poisson rnd variable In this exercise:

Let $X$ be a Poisson random variable with parameter $\lambda$ and let $Y=X+1$. Find $\mathbb{E}(Y)$ and $\mathrm{var}(Y)$.

I was able to apply the definition of expected value and obtain:
$$
\mathbb{E}(Y)=\sum_{k=0}^{\infty}\frac{k+1}{k!}\lambda^ke^{-\lambda}
$$
also, after some manipulation and simplification I was able to get it down to 
\begin{align}
\mathbb{E}(Y)&=\sum_{k=1}^{\infty}\frac{e^{-\lambda}}{(k-1)!}\lambda^k+e^{-\lambda}
\end{align}
but now I'm not sure how to continue to simplify.
 A: You may be expected just to quote the results $E(aX+b)=aE(X)+b$ and $\text{Var}(aX+b)=a^2\text{Var}(X)$, and then use the known mean and variance of the Poisson.
But for the part of the computation that gave you trouble, you want 
$$\sum_{k=0}^\infty k e^{-\lambda}\frac{\lambda^k}{k!}.\tag{1}$$
The term $k=0$ contributes nothing. So our sum (1) is equal to
$$\lambda \sum_{k=1}^\infty e^{-\lambda}\frac{\lambda^{k-1}}{(k-1)!}.\tag{2}$$
Let $i=k-1$. Then (2) is equal to
$$\lambda\sum_{i=0}^\infty  e^{-\lambda}
\frac{\lambda^i}{i!}.\tag{3}$$
We recognize the infinite sum $\sum_{i=0}^\infty  e^{-\lambda}
\frac{\lambda^i}{i!}$ as the sum over all $i$ of the Poisson probabilities. So that sum is $1$, and Expression (3) collapses to $\lambda$.
Alternately, we can write (3) as $\lambda e^{-\lambda}\sum_{i=0}^\infty
\frac{\lambda^i}{i!}$, and recognize the sum as the Maclaurin expansion of $e^{\lambda}$.  
A similar calculation can be used for the variance, if we wish to calculate the variance "from scratch." It is somewhat  more complicated. You will likely end up needing to find, among other things, $\sum_{k=0}^\infty k^2 \frac{\lambda^k}{k!}$. 
