I'm studying Wiener process (Brownian motion) with the book: Brzezniak, & Zastawniak. "Basic Stochastic Processes." Springer.
I don't understand the solution for one of the exercises in the book. I'll first present the preliminary definition written in the book.
Definition Let $$0=t_0^n<t_1^n<\cdots<t_n^n=T,$$ where $$t_i^n={iT\over n},$$ be a partition of the interval $[0, T]$ into $n$ equal parts. We denote by $$\Delta_i^n W = W(t_{i+1}^n) - W(t_i^n)$$ the corresponding increments of the Wiener process $W(t)$.
Now here's the exercise that I'm stuck with.
Exercise 6.29 Show that $$\operatorname{lim}_{n\to\infty}\Sigma_{i=0}^{n-1}(\Delta_i^nW)^2 = T\quad \mathrm{in} \ L^2.$$
Here's the solution from the book.
Soltion Since the increments $\Delta_i^n W$ are independent and $$E(\Delta_i^n W) = 0,\quad E((\Delta_i^n W)^2) = {T \over n}, \quad E((\Delta_i^n W)^4) = {3T^2 \over n^2},$$ it follows that $$E([\Sigma_{i=0}^{n-1}(\Delta_i^n W)^2-T]^2) = E([\Sigma_{i=0}^{n-1}((\Delta_i^nW)^2-{T\over n})]^2)$$
$$=\Sigma_{i=0}^{n-1}E[((\Delta_i^nW)^2-{T\over n})^2]$$ $$=\Sigma_{i=0}^{n-1}[E((\Delta_i^n W)^4) - {2T\over n}E((\Delta_i^n W)^2) + {T^2 \over n^2}]$$ $$=\Sigma_{i=0}^{n-1}[{3T^2\over n^2}-{2T^2\over n^2}+{T^2\over n^2}]={2T^2\over n}\to 0$$ as $n\to\infty$.
I don't get the part $E([\Sigma_{i=0}^{n-1}((\Delta_i^nW)^2-{T\over n})]^2) = \Sigma_{i=0}^{n-1}E[((\Delta_i^nW)^2-{T\over n})^2]$ in the solution. How does a square of a sum becomes a sum of squares?
\begin{align} \mathsf{E}\left[\sum_{i=0}^{n-1}\left((\Delta_i^nW)^2-{T\over n}\right)\right]^2&=\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\mathsf{E}\left((\Delta_i^nW)^2-{T\over n}\right)\left((\Delta_j^nW)^2-{T\over n}\right) \\ &=\sum_{i=0}^{n-1}\mathsf{E}\left((\Delta_i^nW)^2-{T\over n}\right)^2 \\ &\quad+ \sum_{i=0}^{n-1}\sum_{j\ne i}\mathsf{E}\left((\Delta_i^nW)^2-{T\over n}\right)\mathsf{E}\left((\Delta_j^nW)^2-{T\over n}\right) \\ &=\sum_{i=0}^{n-1}\mathsf{E}\left((\Delta_i^nW)^2-{T\over n}\right)^2 \end{align} because the increments of $W$ are independent.