I've spent so much time trying to proof to myself these are equivalent, but it's not working out.
Given some $X = u + Y$,
where $Y \sim N(0,1)$.
I have some estimator $Z = 2x_0 + \sum_{i=0}^N x_i^2 + 2x_{N+1}$
$Var[z] = E[Z^2] = E[(2x_0 + \sum_{i=0}^N x_i^2 + 2x_{N+1})^2] = 4 + N + 4 = N + 8$
because the variance is distributed.
$ Var[2x_0] + Var[\sum_{i=0}^N x_i^2] + Var[2x_{N+1}] $
However, when I try to distribute the terms as a polynomial instead, I definitely don't get that answer. I know I'm doing something wrong somewhere.
$Var[z] = E[Z^2] = E[(2x_0 + \sum_{i=0}^N x_i^2 + 2x_{N+1})^2] = E[4x_0^2 + 2x_0\sum_{i=0}^N x_i^2+...+2x_N\sum_{i=0}^N x_i^2+4x_{N+1}^2]$
Your calculations of the variance are not correct.
You can write $X\sim N(u,1)$. And $X^2 = u^2 + Y^2 + 2uY$. Thus, $$E[X] = u$$ and $$E[X^2] = u^2+1,$$ this is just the second moment of $X$.
Then the variance of $Z$ is defined as \begin{aligned} \text{Var}[Z] &= E[ (Z - E[Z])^2 ]\\ &= E[Z^2] - (E[Z])^2 \end{aligned}
Note that $$ E[Z] = 4u + (N+1)(u^2+1). $$