Expanding $E[N^2]-E[N]$ I'm trying to prove $$\sum_{i=0}^{\infty}iP[N>i]=\frac{1}{2}(E[N^2]-E[N])$$
by expanding both the RHS and LHS and showing that they are equal. The first thing I did was multiply both sides by $2$ to get $$2\sum_{i=0}^{\infty}iP[N>i]=E[N^2]-E[N]$$
which made it simpler to expand the LHS. I ended up getting the following equality:
$$2\sum_{i=0}^{\infty}iP[N>i] = 2P[N=2] + 6P[N=3] + 14P[N=4] + \cdots$$
When trying to expand the LHS, however, I did not know how to proceed once I saw $E[N^2]$. So, I have 2 questions:
(1) Is the following true: $E[N^2]-E[N] = E[N^2 - N] = E[N(N-1)]?$
(2) Can $E[N^2]$ be expanded in a similar way to $E[N]$? i.e. $E[N^2] = 0^2P[N=0] + 1^2P[N=1] + 2^2P[N=2^2]$? I know this is 100% incorrect, but I can't figure out how else it would be expanded.
Thank you!

Yes, what you wrote $\Bbb E[N^2]-\Bbb E[N]=\Bbb E[N(N-1)]$ is correct. You can expand it as $$\Bbb E\left[N(N-1)\right]=\sum_{i=1}^{\infty}i(i-1)P(N=i)=2P(N=2)+6P(N=3)+12P(N=4)+\dots$$ Now, go back and check where this $14P(X=4)$ comes from in the expansion of the LHS. It should be $12P(X=4)$. 

Your expansion of $\Bbb E\left[N^2\right]$ is correct as well.
