= Solution
First take $1\leq n\leq F-1$. The forward object reaches $n$ at its $n$th move, while the reverse object reaches that label at its $(F-n)$th move. If the forward object wins after $k$ reverse moves, then $0\leq k\leq F-n-1$. The final choice is forward; among the preceding $n+k-1$ choices, exactly $k$ are reverse. By the <negative binomial stopping argument>, this case has <probability> $\binom{n-1+k}{k}p^nq^k$.
Sum these mutually exclusive cases and equate the result to part (e), using parts (b) and (d). This proves the <success-failure stopping identity> here:
$$
\sum_{k=0}^{F-n-1}\binom{n-1+k}{k}p^nq^k
=\binom{F-1}{n}p^nq^{F-n}
+\sum_{k=n+1}^{F}\binom Fk p^kq^{F-k}.
$$
The right side splits the same race into a special meeting at $n$ and all meetings beyond $n$; the left side splits it according to the number of reverse moves at the winning time.
At $n=F$, both sides are zero with an <empty sum> and $\binom{F-1}{F}=0$. At $n=0$, the event has <probability> one; the right side equals $\sum_{k=0}^{F}\binom Fk p^kq^{F-k}=1$. The printed left side includes $\binom{-1}{0}$ and is outside the usual nonnegative-upper-index combinatorial definition. With the explicit generalized convention \b[$\binom{-1}{0}=1$], its $k=0$ term is one and all $k\geq1$ terms vanish, so the identity also holds at that endpoint. Without this convention the displayed sum is undefined there; the zero-target race must instead be stated separately. All identities extend to $p=0,1$ by their <polynomial> forms.
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