= Solution
Group the random resistance cost by unoriented <edges>. For $e=\{x,y\}$, its resistance contributes once whenever the walk traverses it in either direction. Pathwise,
$$
\sum_{t=0}^{\tau-1}R_{\mathrm{eff}}(X_t,X_{t+1})=\sum_{\{x,y\}\in E}R_{\mathrm{eff}}(x,y)\bigl(S(x,y)+S(y,x)\bigr).
$$
The <graph> is finite and the commute has finite <expected value>. <Linearity of expectation>, or <Tonelli theorem> for these nonnegative terms, therefore permits taking expectations of the sum. By the <directed edge occupation in a random-walk commute>, both directed traversal counts have <expected value> $R_{\mathrm{eff}}(a,z)$. Thus
$$
\mathbb E_a\sum_{t=0}^{\tau-1}R_{\mathrm{eff}}(X_t,X_{t+1})=2R_{\mathrm{eff}}(a,z)\sum_{e\in E}R_{\mathrm{eff}}(e).
$$
Apply <Foster's theorem> from part (a) to obtain
$$
\boxed{\mathbb E_a\sum_{t=0}^{\tau-1}R_{\mathrm{eff}}(X_t,X_{t+1})=2(n-1)R_{\mathrm{eff}}(a,z).}
$$
The factor two counts the two orientations of each <edge>. It would be lost by treating the expected directed count in part (b) as an expected count for both directions together. The same distinct-terminal and simple-random-walk conventions from part (b) remain in force.
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