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.
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?