In order to prove that the mean duration of the random walk $m\left(x\right)$ is finite for all $x\in\left(A,B\right)$ (e.g. $x$ is the starting position of a particle). For the case $x=0$ the author introduces an equation: $$S_{\tau_{n}}=\sum_{k=0}^{n}S_{k}\left(\omega\right)I_{\left\{ \tau_{n}=k\right\} }\left(\omega\right)$$
$\xi_{1},\xi_{2},\dots,\xi_{n}$ is a sequence of independent Bernoulli random variables.
$P\left(\xi_{i}=1\right)=p$, $P\left(\xi_{i}=-1\right)=q$, $p\neq q$
$S_{0}=0$, $S_{k}=\sum_{i=1}^{k}\xi_{i}$
$\tau_{n}$ is stopping time.
In page 93, equation (34): $$\mathrm{E}\left[S_{\tau_{n}}-\tau_{n}\cdot\mathrm{E}\xi_{1}\right]^{2}=\mathrm{Var}\xi_{1}\cdot\mathrm{E}\tau_{n}$$
which said to be proved similarly to previous method, in the case of $p=q$, introduced in previous page. But, I could not have the result above and stuck: $$\begin{align*} \mathrm{E}\left[S_{\tau_{n}}-\tau_{n}\cdot\mathrm{E}\xi_{1}\right]^{2} & =\mathrm{E}\sum_{k=0}^{n}\left[S_{k}^{2}-2kS_{k}\mathrm{E}\xi_{1}+k^{2}\left(\mathrm{E}\xi_{1}\right)^{2}\right]I_{\left\{ \tau_{n}=k\right\} }\\ & =\mathrm{E}\sum_{k=0}^{n}\left[\left(S_{k}-S_{n}+S_{n}\right)^{2}-2kS_{k}\mathrm{E}\xi_{1}+k^{2}\left(\mathrm{E}\xi_{1}\right)^{2}\right]I_{\left\{ \tau_{n}=k\right\} }\\ & =\mathrm{E}S_{n}^{2}+\mathrm{E}\sum_{k=0}^{n}\left[2S_{n}\left(S_{k}-S_{n}\right)+\left(S_{n}-S_{k}\right)^{2}-2kS_{k}\mathrm{E}\xi_{1}+k^{2}\left(\mathrm{E}\xi_{1}\right)^{2}\right]I_{\left\{ \tau_{n}=k\right\} }\\ & =\mathrm{E}S_{n}^{2}-\sum_{k=0}^{n}\mathrm{E}\left(S_{n}-S_{k}\right)^{2}I_{\left\{ \tau_{n}=k\right\} }+\mathrm{E}\sum_{k=0}^{n}\left[-2kS_{k}\mathrm{E}\xi_{1}+k^{2}\left(\mathrm{E}\xi_{1}\right)^{2}\right]I_{\left\{ \tau_{n}=k\right\} }\\ & =\mathrm{E}S_{n}^{2}-\sum_{k=0}^{n}\mathrm{E}\left(S_{n}-S_{k}\right)^{2}\mathrm{E}I_{\left\{ \tau_{n}=k\right\} }+\mathrm{E}\sum_{k=0}^{n}\left[-2kS_{k}\mathrm{E}\xi_{1}+k^{2}\left(\mathrm{E}\xi_{1}\right)^{2}\right]I_{\left\{ \tau_{n}=k\right\} }\\ & =\mathrm{E}S_{n}^{2}-\sum_{k=0}^{n}\left[\left(n-k\right)\mathrm{Var}\xi_{1}+\left(n-k\right)^{2}\left(\mathrm{E}\xi_{1}\right)^{2}\right]\mathrm{E}I_{\left\{ \tau_{n}=k\right\} }+\mathrm{E}\sum_{k=0}^{n}\left[-2kS_{k}\mathrm{E}\xi_{1}+k^{2}\left(\mathrm{E}\xi_{1}\right)^{2}\right]I_{\left\{ \tau_{n}=k\right\} }\\ & =\mathrm{E}S_{n}^{2}-n\mathrm{Var}\xi_{1}+\mathrm{Var}\xi_{1}\mathrm{E}\tau_{n}-\sum_{k=0}^{n}\left(n^{2}-2nk+k^{2}\right)\left(\mathrm{E}\xi_{1}\right)^{2}P\left(\tau_{n}=k\right)+\mathrm{E}\sum_{k=0}^{n}\left[-2kS_{k}\mathrm{E}\xi_{1}+k^{2}\left(\mathrm{E}\xi_{1}\right)^{2}\right]I_{\left\{ \tau_{n}=k\right\} }\\ & =\left(\mathrm{E}S_{n}^{2}-\mathrm{Var}S_{n}\right)+\mathrm{Var}\xi_{1}\mathrm{E}\tau_{n}-\sum_{k=0}^{n}\left(n^{2}-2nk+k^{2}\right)\left(\mathrm{E}\xi_{1}\right)^{2}P\left(\tau_{n}=k\right)+\mathrm{E}\sum_{k=0}^{n}\left[-2kS_{k}\mathrm{E}\xi_{1}+k^{2}\left(\mathrm{E}\xi_{1}\right)^{2}\right]I_{\left\{ \tau_{n}=k\right\} }\\ & =\left(\mathrm{E}S_{n}\right)^{2}+\mathrm{Var}\xi_{1}\mathrm{E}\tau_{n}-n^{2}\left(\mathrm{E}\xi_{1}\right)^{2}+2n\left(\mathrm{E}\xi_{1}\right)^{2}\mathrm{E}\tau_{n}-\left(\mathrm{E}\xi_{1}\right)^{2}\sum_{k=0}^{n}k^{2}P\left(\tau_{n}=k\right)+\left(\mathrm{E}\xi_{1}\right)^{2}\mathrm{E}\sum_{k=0}^{n}k^{2}I_{\left\{ \tau_{n}=k\right\} }-2\mathrm{E}\xi_{1}\mathrm{E}\sum_{k=0}^{n}kS_{k}I_{\left\{ \tau_{n}=k\right\} }\\ & =\left(\mathrm{E}S_{n}\right)^{2}+\mathrm{Var}\xi_{1}\mathrm{E}\tau_{n}-n^{2}\left(\mathrm{E}\xi_{1}\right)^{2}+2n\left(\mathrm{E}\xi_{1}\right)^{2}\mathrm{E}\tau_{n}-2\mathrm{E}\xi_{1}\mathrm{E}\sum_{k=0}^{n}kS_{k}I_{\left\{ \tau_{n}=k\right\} }\\ & =n^{2}\left(\mathrm{E}\xi_{1}\right)^{2}+\mathrm{Var}\xi_{1}\mathrm{E}\tau_{n}-n^{2}\left(\mathrm{E}\xi_{1}\right)^{2}+2n\left(\mathrm{E}\xi_{1}\right)^{2}\mathrm{E}\tau_{n}-2\mathrm{E}\xi_{1}\mathrm{E}\sum_{k=0}^{n}kS_{k}I_{\left\{ \tau_{n}=k\right\} }\\ & =\mathrm{Var}\xi_{1}\mathrm{E}\tau_{n}+2n\left(\mathrm{E}\xi_{1}\right)^{2}\mathrm{E}\tau_{n}-2\mathrm{E}\xi_{1}\mathrm{E}\sum_{k=0}^{n}kS_{k}I_{\left\{ \tau_{n}=k\right\} }\\ \end{align*}$$