Question 3.14: Independent trials, each of which is a success with probabil......

Let $N_{k}$ denote the number of necessary trials to obtain k consecutive successes, and let $M_{k}$ denote its mean. We will obtain a recursive equation for $M_k$ by conditioning on $N_{k−1,}$ the number of trials needed for k −1 consecutive successes. This yields

$M_k=E\left[N_k\right]=E\left[E\left[N_k \mid N_{k-1}\right]\right]$
Now,
$E\left[N_k \mid N_{k-1}\right]=N_{k-1}+1+(1-p) E\left[N_k\right]$
where the preceding follows since if it takes $N_{k-1}$ trials to obtain $k-1$consecutive successes, then either the next trial is a success and we have our k in a row or it is a failure and we must begin anew. Taking expectations of both sides of the preceding yields
$M_k=M_{k-1}+1+(1-p) M_k$
or
$M_k=\frac{1}{p}+\frac{M_{k-1}}{p}$
Since $N_1$, the time of the first success, is geometric with parameter p, we see that
$M_1=\frac{1}{p}$
and, recursively
$\begin{aligned}& M_2=\frac{1}{p}+\frac{1}{p^2}, \\& M_3=\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}\end{aligned}$
and, in general,