I'm trying to find the $\mathbb{E}[X(X+1)]$ for the geometric distribution.
Everywhere I've looked explains how to do it using the derivative but I have not been taught that method. Is there another way to do this? I know that $\mathbb{E}[X(X+1)] = E[X^2]+\mathbb{E}[X]$, but I am unable to calculate $E[X^2]$. Any tips would be greatly appreciated. Thank you in advance!
Note that $$\mathbb{E}[X(X+1)]=\mathbb{E}[X^{2}+X]=\mathbb{E}[X^{2}]+\mathbb{E}[X]$$ Now, we know that $$\mathbb{V}ar[X]=\mathbb{E}[X^{2}]-[\mathbb{E}[X]]^{2}$$ Using that, we can see that $$\mathbb{E}[X(X+1)]=\mathbb{V}ar[X]+[\mathbb{E}[X]]^{2}+\mathbb{E}[X]$$ So, since that $X\sim \textbf{Geometric}(p)$ so, $$\mathbb{E}[X]=\frac{1}{p} \quad \text{and} \quad \mathbb{V}ar[X]=\frac{1-p}{p^{2}}$$
Other approach is by definition, we can see that $$\mathbb{E}[X(X+1)]=\mathbb{E}[X^{2}+X]=\mathbb{E}[X^{2}]+\mathbb{E}[X]$$ Now, we know that $\mathbb{E}X=\frac{1}{p}$ and we can see that $$\mathbb{E}[X^{2}]=\sum_{x}x^{2}\mathbb{P}[X=x]$$ with $x=1,2,\ldots$ and $\mathbb{P}[X=x]=p(1-p)^{x-1}$. So, you need to find $$\mathbb{E}[X^{2}]=\sum_{x}x^{2}\mathbb{P}[X=x]=\sum_{x=1}^{\infty}x^{2}p(1-p)^{x-1}$$ Using that $0<p\leq 1$, so we can find that $$\mathbb{E}[X^{2}]=\sum_{x}x^{2}\mathbb{P}[X=x]=\sum_{x=1}^{\infty}x^{2}p(1-p)^{x-1}=\frac{2-p}{p^{2}}$$