Mean squared error of the dyadic quadratic variation

Let $B$ be standard Brownian motion and fix a horizon $t>0$. Cut $[0,t]$ into $2^n$ intervals of equal length and write $D_k=B_{kt/2^n}-B_{(k-1)t/2^n}$, $k=1,\dots,2^n$. Derive a formula, in terms of $t$ and $n$, for

then evaluate it for $n=5$ and $t=1$.

$2t^2/2^n$, which is $1/16$ at $t=1$, $n=5$. The sum of squared increments has mean exactly $t$, so the quantity is its variance. The increments are independent $N(0,t/2^n)$, each square has variance $2(t/2^n)^2$, and adding $2^n$ of them gives $2t^2/2^n=2/32$.