# Hitting time on linear Markov chain

I came across this problem in a math course of mine a while back, and I haven't been able to solve it since. Anyone have any ideas?

Suppose we have a chain of $$n$$ vertices, such that the first vertex has an edge to the second, the $$i$$th vertex has an edge to the $$(i-1)$$th and $$(i+1)$$th vertices, and the $$n$$th vertex has an edge to only the $$(n-1)$$th vertex. If we perform a random walk on the graph, starting at the second vertex, what is $$E(h)$$, if $$h$$ is the number of steps it takes to reach the first vertex?

I wrote a program that performs ~100,000 random walks to find a reasonable answer, and from that, I was able to determine that it takes $$O(n)$$ time, but I have not been able to find any insight into the combinatorial nature of the problem that gives us this result.

How do I prove this result rigorously?

Furthermore, what is the expected hitting time $$E(h_i)$$ when performing a random walk starting at vertex $$i$$?

Note: Another result that I know from the textbook is that $$E(h_n) = n^2$$, but again, I have not yet been able to come up with a proof of this.