How do you go from the first step where the summation of ( 1 − p ) k − 1 transform to the fraction ∑ k = 1 n p ( 1 − p ) k − 1 = p 1 − ( 1 − p ) n 1 − ( 1 − p ) How to simplify the summation?

Question

How do you go from the first step where the summation of   (  1  −  p      )          k      −      1       transform to the fraction      ∑          k      =      1        n    p  (  1  −  p      )          k      −      1        =  p            1      −      (      1      −      p              )        n                    1      −      (      1      −      p      )      How to simplify the summation?

Teagan Zamora · Accepted Answer

Explanation:For all   a  ≠  1, one can prove that       ∑          k      =      0        n        a    k    =            1      −              a                  n          +          1                            1      −      a        , for all   n  ∈      N

djo57bgfrqn · Accepted Answer

Step 1Firstly, lets get rid of p,which appears everywhere:      ∑          k      =      1              n        p  (  1  −  p      )          k      −      1        =  p      ∑          k      =      1              n        (  1  −  p      )          k      −      1      Now,lets denote   a  =  1  −  p,where a is an integer.Now denote with   S  =      ∑          k      =      1              n            a          k      −      1      Lets calculate the product between a and S:  a  S  =      ∑          k      =      1              n            a          k        a  S  −  S  =      ∑          k      =      1              n            a          k        −  SNow lets subtract S from the sum:  (  a  −  1  )  S  =      ∑          k      =      1              n            a          k        −      ∑          k      =      1              n            a          k      −      1        (  a  −  1  )  S  =      a    n    −  1  S  =                    a        n            −      1              a      −      1        S  =            1      −              a        n                    1      −      a      Step 2Now all that we have to do is to replace a with   p  −  1  S  =            1      −      (      1      −      p              )        n                    1      −      (      1      −      p      )        p  S  =  p            1      −      (      1      −      p              )        n                    1      −      (      1      −      p      )        p            1      −      (      1      −      p              )        n                    1      −      (      1      −      p      )        =  p  S  =  p      ∑          k      =      1              n        (  1  −  p      )          k      −      1        =      ∑          k      =      1              n        p  (  1  −  p      )          k      −      1

How do you go from the first step where the summation of (1-p)^{k-1} transform to the fraction

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