= Solution
To obtain the $(k+1)$st head on toss $n+1$, the first $n$ tosses must contain exactly $k$ heads, and the last toss must be a head. There are $\binom nk$ choices of the first $k$ head positions. By <independence>, each such arrangement followed by a head has <probability> $r^k(1-r)^{n-k}r$. The arrangements are disjoint, so
$$
\boxed{h_{n,k}(r)=\binom nk(1-r)^{n-k}r^{k+1},\qquad n\geq k\geq0.}
$$
This is the <negative binomial stopping argument>, with $n+1$ counting all tosses up to and including the $(k+1)$st success. It is not merely the <probability> of $k+1$ heads somewhere among $n+1$ tosses: the last toss is prescribed to be a head.
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