How to find sum of infinite geometric series with coefficient?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. ∑ n = 0 ∞ p ( 1 − p ) 3 n 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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How to find sum of infinite geometric series with coefficient?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&#039;s probability.      ∑          n      =      0              ∞        p  (  1  −  p      )          3      n      Can you use geometric series sum formula? Is it   p      /    (  1  −  (  1  −  p      )    3    )?How do you deal with 3 that&#039;s in front of n?

Hassan Watkins · Accepted Answer

Step 1Your sum is correct. All you need to do is recognize that       ∑          n      =      0              ∞        p  (  1  −  p      )          3      n        =      ∑          n      =      0              ∞        p            [      (      1      −      p              )        3            ]        n  Step 2The latter series is a geometric series whose common ratio and leading term are   (  1  −  p      )    3   and p, respectively, so it will converge to       p          1      −      (      1      −      p              )        3             if       |    (    1    −    p          )      3        |    &amp;lt;  1 and diverge if       |    (    1    −    p          )      3        |    ≥  1

How to find sum of infinite geometric series with coefficient? Given this series, p+p(1-p)^3+p(1-p)^6+p(1-p)^9+...

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