Suppose $a_k=\frac{1}{2}p(a_{k+1}+1)+\frac{1}{2}p(a_{k-1}+1)+(1-p)(a_k+1)$ for $0<k<n$ and $a_0=a_n=0$ and $0<p<1$. How do we find a closed form of $a_k$? For $p=1$, I know the solution is $a_k=k(n-k)$. But I have no idea about this case. It is a system of linear equations, so I think it will a unique solution. In fact, the problem comes from random walk on $\{0,1,2,...,n\}$ with absorbing states $0,n$. And $a_k$ is the expectation of number of steps reaching absorbing states when starting at $k$.
$a_k=\frac{1}{2}p(a_{k+1}+1)+\frac{1}{2}p(a_{k-1}+1)+(1-p)(a_k+1)\Leftrightarrow pa_k=1+\frac{1}{2}pa_{k-1}+\frac{1}{2}pa_{k+1}$. Thus if we let $b_k=pa_k,$ then it is equivalent to $b_k=1+\frac{1}{2}b_{k-1}+\frac{1}{2}b_{k+1}$, which has solution as above. But how to solve this one about $b_k$?
Hint: sum $\;a_{j+1} - 2 a_j + a_{j-1} = -\frac{2}{p}\;$ for $\,j=1,\cdots,k\,$ and watch most terms telescope, leaving $a_{k+1}-a_{k}-a_{1}+a_{0}=-\frac{2k}{p}\implies a_{k+1}-a_{k}=a_1-\frac{2k}{p}$ since $a_0=0\,$.
Sum $\;a_{j+1}-a_{j}=a_1-\frac{2j}{p}\;$ for $\,j=1,\cdots,k\,$ and telescope again: $\require{cancel}\,a_{k+1}-a_1=k a_1 - \frac{\cancel{2}}{p} \frac{k(k+1)}{\cancel{2}}\,$, or $\;a_{k}=k a_1 - \frac{k(k-1)}{p}\,$.
Lastly, let $k=n-1$ and determine $a_1$ from the condition $0 = a_n = n a_1 - \frac{n(n-1)}{p}\,$, which gives $a_1=\frac{n-1}{p}\,$, so in the end $a_k= k a_1- \frac{k(k-1)}{p}=\frac{k(n-1)}{p}- \frac{k(k-1)}{p}=\frac{k(n-k)}{p}\,$.