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

### 1 Answer | Add Yours

**`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)` **

**Sources:**