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If `A_n = 1-(1-beta)alpha^(n-1)` Find `sum _(r=1)^n A_r`

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roshan-rox

Posted September 22, 2013 at 1:16 AM via web

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If `A_n = 1-(1-beta)alpha^(n-1)`

Find `sum _(r=1)^n A_r`

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted September 22, 2013 at 1:30 AM (Answer #1)

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`sum _ (r=1)^n A_ r = sum _ (r=1)^n [1-(1-beta)alpha^(r-1)]`

`=sum _ (r=1)^n 1-(1-beta) sum _ (r=1)^n alpha^(r-1)`

`sum_(r = 1)^n(alpha)^(r-1)` represent the sum of a geometric series.

a = 1

`r = alpha`

Also `sum_(r = 1)^nA_r = 1+1+1.....+1 = n`

`sum _ (r=1)^n A_ r = sum _ (r=1)^n [1-(1-beta)alpha^(r-1)]`

`=n-(1-beta)(1-alpha^n)/(1-alpha)`

So the answer is;

`sum _ (r=1)^n A_ r = n-((1-beta)(1-(alpha)^n))/(1-alpha)`

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