Solution
= Solution
Measure vertices by their <Hamming distance> from the target and let $h_k$ be the expected hitting time when that distance is $k$. We already know $h_1=2^n-1$. From distance one, one of the $n$ coordinate flips reaches the target and the other $n-1$ flips move to distance two. <First-step analysis> therefore gives
$$
h_1=1+\frac1n h_0+\frac{n-1}{n}h_2,
\qquad h_0=0.
$$
For $n\geq2$,
$$
\boxed{h_2=\frac{n}{n-1}(h_1-1)
=\frac{n(2^n-2)}{n-1}.}
$$