I am trying to solve part (a) of this exercise, which states that
$$E\left(\left[\sum_{t_k\leq t} \left(\Delta B_k\right)^2 - t\right]^2 \right) =2\sum_{t_k\leq t}(\Delta t_k)^2 ~, $$
where $0=t_0<\dots<t_n=t$. What I've done so far is squaring in the LHS and taking the expectancy, so I get
$$ \sum E\left(\left(\Delta B_k\right)^4\right)-t^2 +t^2 $$
I'm stuck at computing the expectancy that is left, this is probably elementary but I haven't succeeded.
I arrived at the following solution using @saz's suggestion. First we square $$E[(\sum_{t_k\leq t} (\Delta B_k)^2 -t)^2]=\sum_{t_k\leq t}E((\Delta B_k)^4) - t \sum_{t_k\leq t}E((\Delta B_k)^2)+t^2$$ here we are using the fact that the increments are independent. We have that $$Var((\Delta B_k)^2)=E((\Delta B_k)^4)-E((\Delta B_k)^2)^2=E((\Delta B_k)^4)-(\Delta t_k)^2$$ And using the fact that $(\Delta B_k)^2$ has a $\chi^2$ distribution with 1 degree of freedom, we get $Var((\Delta B_k)^2)=2(\Delta t_k)^2$. So $E((\Delta B_k)^4)=(\Delta t_k)^2$. So now we have $$E[(\sum_{t_k\leq t} (\Delta B_k)^2 -t)^2]=\sum_{t_k\leq t}(\Delta t_k)^2 - t \sum_{t_k\leq t}\Delta t_k +t^2=\sum_{t_k\leq t} (\Delta t_k)^2$$ Since the second is a telescoping sum. I might be mistaken at some point since the 2 doesn't appear, but maybe it's a typo?