Given $\left\{A_n\right\}$ be a zero mean and $F_n$- martingale with $E(A_n^2)<\infty$. Prove that $$E(A_{n+r}-A_n)^2 = \sum_{k=1}^{r} E(A_{n+k}-A_{n+k-1})^2$$
My attempt: We rewrite the LHS as: \begin{align}E(A_{n+r} - A_n)^2= &E(A_{n+r} - A_{n+r-1})^2 + E(A_{n+r} - A_{n+r-1})^2 \\[0.2cm]&+ 2E(A_{n+r} - A_{n+r-1})(A_{n+r-1}-A_n) + E(A_{n+r-1}-A_n)^2\end{align} Now, we will show that $E(A_{n+r} - A_{n+r-1})(A_{n+r-1}-A_n) = 0$ as follows: \begin{align}E(A_{n+r}A_{n+r-1})& = EE(A_{n+r}A_{n+r-1}\mid F_{n+r-1}) = E(A_{n+r-1}E(A_{n+r}\mid F_{n+r-1})) \\[0.2cm]&= E(A_{n+r-1}A_{n+r-1})\end{align}
\begin{align}E(A_{n+r-1}A_r) - E(A_{n+r}A_r) &= EE(A_{n+r-1}-A_{n+r})A_r\mid F_{n+r-1})\\[0.2cm]& = E((A_{n+r-1}-A_{n+r})E(A_r\mid F_{n+r-1})) = 0\end{align} since $E(A_r\mid F_{n+r-1}) =$ some constants, and $E((A_{n+r-1}-A_{n+r}) = 0-0=0$. Now repeat the whole process above for $E(A_{n+r-1}-A_n)^2$ and the following terms, we achieve what we want.
My question: Could anyone please help verify if my solution above is correct? I am quite skeptical about the part of proving $$E(A_{n+r-1}A_r) - E(A_{n+r}A_r) = 0$$ So if someone could help, I would really appreciate.
Your proof looks okay. I would like to show you the following alternative reasoning (which seems a little bit easier to me).
Due to the tower property of conditional expectation and the martingale property, we have
$$\mathbb{E}(A_{m+r} A_m) = \mathbb{E} \bigg[ A_m \mathbb{E}(A_{m+r} \mid F_m) \bigg] = \mathbb{E}(A_m^2) \tag{1}$$
for all $r \in \mathbb{N}$ and $m \in \mathbb{N}$. Hence,
$$\begin{align*} \mathbb{E}((A_{m+r}-A_m)^2) &= \mathbb{E}(A_{m+r}^2)- 2 \mathbb{E}(A_{m+r} A_m) + \mathbb{E}(A_m^2) = \mathbb{E}(A_{m+r}^2)-\mathbb{E}(A_m^2). \tag{2} \end{align*}$$
Since this holds for all $r \geq 1$ and $m \in \mathbb{N}$, we get
$$\begin{align*} \mathbb{E}((A_{n+r}-A_n)^2) \stackrel{(2)}{=} \mathbb{E}(A_{n+r}^2)-\mathbb{E}(A_n^2) &= \sum_{k=1}^r (\mathbb{E}(A_{n+k}^2)- \mathbb{E}(A_{n+k-1}^2)) \\ &\stackrel{(2)}{=} \sum_{k=1}^r \mathbb{E}((A_{n+k}-A_{n+k-1})^2) \end{align*}$$ where we have used in the last step $(2)$ with $r=1$ and $m=n+k-1$.