Prove the general case of Bienaymé's Formula: if $X_i$, $i=1,2,...,n$, are pairwise independent random variables on a sample space $S$, then
$$V(X_1 + X_2 + ... + X_n) = V(X_1) + V(X_2) + ... + V(X_n)$$
using the fact that $$(\sum_{i=1}^n a_i)^2 = \sum_{i=1}^n a_i^2 + 2\sum_{1 \le i \lt j \le n}a_ia_j$$
I am able to work out the proof for the case with two random independent variables as follows: $$V(X+Y) = E((X+Y)^2) - E(X+Y)^2$$ $$=E(X^2 + 2XY + Y^2) - (E(X) + E(Y))^2$$ $$=E(X^2) + 2E(XY) + E(Y^2) - E(X)^2-2E(X)E(Y) - E(Y^2)$$ $$=E(X^2) -E(X)^2 + E(Y^2) - E(Y^2)$$ $$=V(X) + V(Y)$$
Following that model, I have $$V(\sum_{i=1}^n X_i)$$ $$=E((\sum_{i=1}^n X_i)^2) - (E(\sum_{i=1}^n X_i))^2$$ $$=E(\sum_{i=1}^n X_i^2 + 2\sum_{1 \le i \lt j \le n}X_iX_j) - (E(\sum_{i=1}^n X_i))^2$$
But that's where I get stuck on how to proceed with the second term.
For example you can do like this. Let $X_{1}+X_{2}+ ... + X_{n-1}= Y$, which is independent of $X_{n}$, as it is completely defined by the means of $X_{1} , ... , X_{n-1}$. So as you have for two. Than you'll have $Var(X_{1}+ ... +X_{n-1}+ X_{n}) = Var(Y+X_{n})= Var(Y)+Var(X_{n})$, and so on.