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Given this series,

$ p + p(1-p)^3 + p(1-p)^6 + p(1-p)^9 + ...$

This is an infinite geometric series with ratio less than 1 since it's probability.

$$ \sum_{n=0}^{\infty}p(1-p)^{3n}$$

Can you use geometric series sum formula? Is it $ p / (1-(1-p)^3)$?

How do you deal with 3 that's in front of n?

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    $\begingroup$ Use the substitution $q = (1-p)^3$. $\endgroup$ – Andy Walls Sep 30 '20 at 3:20
  • $\begingroup$ Right but then n is NOT raised to q right? (1-p)^(3n) does not equal (1-p)^3^n $\endgroup$ – Jake Yoon Sep 30 '20 at 3:27
  • $\begingroup$ $(1-p)^{3n}=((1-p)^3)^n$ $\endgroup$ – J. W. Tanner Sep 30 '20 at 3:44
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Your sum is correct. All you need to do is recognize that

$$\sum_{n=0}^{\infty}p(1-p)^{3n}=\sum_{n=0}^{\infty}p\left[(1-p)^3\right]^n$$

The latter series is a geometric series whose common ratio and leading term are $(1-p)^3$ and $p$, respectively, so it will converge to $\frac{p}{1-(1-p)^3}$ if $\left|(1-p)^3\right|<1$ and diverge if $\left|(1-p)^3\right|\geq 1$.

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