= Solution
The <absorbing states> give $e_0=e_n=0$. Multiplying the <difference equation> in (i) by $3$ gives
$$
e_{k+1}-2e_k+e_{k-1}=-3.
$$
The second <finite difference> of $k^2$ is $2$, so a <particular solution> is $-3k^2/2$; the homogeneous solutions are $A+Bk$. Imposing the two <boundary conditions> gives $A=0,B=3n/2$. Therefore
$$
\boxed{e_k=\frac32k(n-k),\qquad0\le k\le n.}
$$
Uniqueness follows because the difference of any two solutions has zero second <finite difference>, hence is <affine function>, and vanishes at both endpoints. This is the <mean absorption time of a lazy symmetric random walk> with holding probability $1/3$.
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