I've got difficulties in computing the second momentum of chi squared.
Chi squared distribution with $n$ degrees of freedom is the sum of $n$ independent distributions $X^2$, where $X \sim N(0;1)$.
We know that the fourth momentum of each $X_i$ is $3$.
So, mark $\chi^2 = C$, and $$E(C^2) = E\left[ \left(\sum X^2\right)^2 \right] = \sum E\left[X^4\right] = 3n$$ where the penultimate equality is due to $X$s being iid and linearity of E-operator.
The correct answer is $n^2 +2n$, so there is something wrong with the above calculation, but I just can't spot what it is. (Please note that I'm not claiming that $\left(\sum x^2 \right)^ 2 = \sum x^4$ in general. Here it should hold however, since the $X$s are independent, right?)