= Solution
Let $\tau$ be the <absorption time>. Its <expectation> is finite: from every interior state, a string of at most $n$ successive left moves reaches $0$, and the <probability> of such a string is at least $p=3^{-n}$. Hence the <Markov property> gives $\mathbb P_k(\tau>jn)\le(1-p)^j$, a geometric tail bound, and $\mathbb E_k[\tau]<\infty$.
For $1\le k\le n-1$, one step is always spent before absorption or the remaining walk. Apply <first-step analysis> and the <law of total expectation> to the three possible first moves. The <conditional expectations> of the remaining <absorption time> are $e_{k+1},e_k,e_{k-1}$, respectively; independence of future moves gives the same law as a fresh <random walk> from the new position. Thus
$$
e_k=1+\frac13e_{k+1}+\frac13e_k+\frac13e_{k-1},
$$
which rearranges to
$$
\boxed{\frac13e_{k+1}-\frac23e_k+\frac13e_{k-1}=-1.}
$$
The term $1$ also appears when the walk stays at $k$, because staying still consumes a step.
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