= Solution
Let $T_k$ be the expected capture time starting from $D_0=k$, so $T_0=0$. The <Strong Markov property> at successive first passages to lower levels and part (a) give
$$
T_k=T_1+(k-1)\mu,
\qquad k\ge1.
$$
At $D_n=1$, the next state is zero with probability $r/2$, one with probability $1/2$, and two with probability $(1-r)/2$. <First-step analysis> therefore gives
$$
T_1=1+\frac12T_1+\frac{1-r}{2}T_2,
$$
or
$$
T_1=2+(1-r)(T_1+\mu).
$$
Solving,
$$
T_1=\frac2r+\left(\frac1r-1\right)\mu.
$$
Substituting $\mu=2/(2q-1)$ now yields
$$
\boxed{
T_m=\frac2r+\left(m+\frac1r-2\right)\frac2{2q-1}},
$$
which is the required expected capture time.
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