If X ~ Poisson(λ), then find E((X+1)^2) I've tried to simplify using the infinite summation but to no avail. Any help would be greatly appreciated.

The mean and variance of a Poisson distribution with parameter $\lambda$ are both $\lambda$.
$$\mathbb{E}[X]=\lambda \text{ and } \mathbb{Var}[X]=\lambda.$$
Since $\mathbb{Var}[X]=\mathbb{E}[X^2]-(\mathbb{E}[X])^2$, we can calculate the expected value of $X^2$:
$$\mathbb{E}[X^2]=\mathbb{Var}[X]+(\mathbb{E}[X])^2=
\lambda+\lambda^2.$$
Expectation is linear, so we have:
$$\begin{aligned}\mathbb{E}[(X+1)^2]&=\mathbb{E}[X^2+2X+1]\\
&=\mathbb{E}[X^2]+2\mathbb{E}[X]+\mathbb{E}[1]\\
&=\lambda+\lambda^2+2\lambda+1\\ &=\lambda^2+3\lambda+1.\end{aligned}$$
