Let $z=2W$, where $W$ is a Brownian Motion.
For $\{t_0^n,t_i^n,...\}=D^n$ (enumeration of dyadic partition), it is possible to show that the processes $$z_t^{(n)}=\sum_{i\geq0}2W_{t_i^n}\mathbb{1}_{]t_i^n,t_{i+1}^n]}(t)$$ converge to $W$ in $\mathscr{H}^2$, where $$\mathscr{H}^2:=\{z | z \text{ measurable adapted stochastic process },E\int_0^T|z_t|^2dt\lt\infty\}$$
We have $$\int_0^tz_s^{(n)}dW_s$$ $$=\sum_{i\ge0}(2W_{t_i^n}(W_{t\wedge{t_{i+1}^n}}-W_{t\wedge{t_{i}^n}}))$$ $$=\sum_{i\ge1}(W_{t\wedge{t_{i+1}^n}}^2-W_{t\wedge{t_{i}^n}}^2)-\sum_{i\ge1}(W_{t\wedge{t_{i+1}^n}}-W_{t\wedge{t_{i}^n}})^2$$ $$=W_t^2-QV_t^{(n)}(W)\rightarrow{W_t^2-t}$$ where the convergence is in $L^2$ and $QV$ stands for Quadratic Variation.
Question: I get stuck on the second equality of the computation, how do we come to the third line. If i have calculated correctly, the second line is equal to$$\sum_{i\ge1}(2W_{t\wedge{t_{i+1}^n}}W_{t\wedge{t_i^n}}-2W_{t\wedge{t_i^n}}^2)$$ and third line $$W_t^2-\sum_{i\ge1}|W_{t\wedge{t_{i}^n}}-W_{t\wedge{t_{i-1}^n}}|^2$$ for me it's not clear why these two parts are equal.
I would appreciate it if anyone could help me understand it. Thanks in advance!