I am working my way through "Random Walks and Random Environments Vol.1" by Barry Hughes. I am trying to derive the second moment of the time until a random walker is absorbed given that it starts at position $l$ ($\left <T(l)^{2}\right >$). Below is the derivation for the first moment which I am including since I think it will clarify the problem, which is that I am trying to use the same approach to derive the second moment.
On pg. 114 the following derivation for the first moment of the time until a random walker is absorbed at step $n$ by hitting either position $0,N$ is derived as follows. The probability of absorption at time $T(l)$ starting from position $l$ ($0<l<N$) is
$P[T(l)=n] = p P[T(l+1)=n-1] + q P[T(l-1)=n-1]$
where $p$ is the probability of a step to the right and $q$ is a step to the left ($p+q=1$).
The expected value of $T(l)$ is
$\left <T(l)\right > = \sum_{n=0}^{\infty} n P[T(l)=n]$
By plugging in the expression for $P[T(l)=n]$ and shifting $n \rightarrow n+1$, the following difference equation can be obtained
$\left <T(l)\right > = p \left <T(l+1)\right > + q\left <T(l-1)\right > +1$
The general solution of this difference equation can be solved as a homogeneous second-order difference equation $\alpha(l) = p \alpha(l+1) + q \alpha(l-1)$ assuming the candidate solution $\alpha(l) = \alpha^{l}$, from which we obtain the solutions $\alpha = 1, q/p$. We are interested in the unbiased walk scenario $p=q$, so there is only one general solution $\alpha = 1$, from which we obtain $\left <T(l)\right >_{general} = A$
For the particular solution we assume the solution $\left <T(l)\right >_{particular} = Bl^{2} + Cl + D$. Plugging this into both sides of our equation for $\left <T(l)\right >$, we obtain $B=-1$, giving us
$\left <T(l)\right > = \left <T(l)\right >_{general} + \left <T(l)\right >_{particular} = A + Cl -l^{2}$
By imposing the boundary conditions
$\left <T(0)\right > = \left <T(N)\right > = 0$
we get
$\left <T(l)\right > = l(N-1)$
Now, I am trying to derive the second moment $\left <T(l)^{2}\right >$ using the above approach. I believe I am close, but am not getting the correct response. First, I define the sum for the second moment and repeat the $n \rightarrow n+1$ shift.
$\left <T(l)^{2}\right > = \sum_{n=0}^{\infty} n^{2} P[T(l)=n] = \sum_{n=1}^{\infty} (n+1)^{2} P[T(l)=n+1] = \sum_{n=1}^{\infty} n^{2} P[T(l)=n+1] + \sum_{n=1}^{\infty} 2n P[T(l)=n+1] + 1$
Giving us
$\left <T(l)^{2}\right > = p \left <T(l+1)^{2}\right > + q \left <T(l-1)^{2}\right > + 2 p \left <T(l+1)\right > + 2 q \left <T(l-1)\right > + 1$
Plugging in our definition of the first moment and setting $p=q$ we obtain
$\left <T(l)^{2}\right > = p \left <T(l+1)^{2}\right > + q \left <T(l-1)^{2}\right > + 2 \left <T(l)\right > - 1 = \frac{1}{2} \left ( \left <T(l+1)^{2}\right > + \left <T(l-1)^{2}\right > \right ) + 2 l(N-1)- 1$
I think the same general solution applies, resulting in $\left <T(l)^{2}\right >_{general} = A$. For the particular solution, I'm unsure of the appropriate form. I started by assuming $\left <T(l)^{2}\right >_{particular} = Bl^{4} + Cl^{3} + Dl^{2} + El + F $ and plugging it into the equation for $\left <T(l)^{2}\right >_{particular}$ and using the same boundary conditions. But this form result not seem to return the correct solution, which is
$\left <T(l)^{2}\right > = \frac{1}{3} l (N-l)(N^{2} + l(N-l) -2) $
Is this the correct approach to derive the second moment? Is it simply a matter of algebra at this point or have I made a mistake.
