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Math Help - Summation of Arithmetico-Geometrical sequence........

  1. #1
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    Summation of Arithmetico-Geometrical sequence........

    This Summation problem has been troubling me for days........Please help!

    Sum to n terms the following series.........

    (1/1) + (2/3) + (3/9) + (4/27) + ............................. to n terms .
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  2. #2
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    Hello, Arka!

    \displaystyle \text{Simplify: }\;\frac{1}{1} + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \frac{5}{3^4} + \hdots + \frac{n}{3^{n-1}}

    \displaystyle \text{We are given: }\quad\; S \;=\;\frac{1}{1} + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \frac{5}{3^4} + \hdots + \frac{n}{3^{n-1}}

    \displaystyle \text{Multiply by }\tfrac{1}{3}:\;\frac{1}{3}S \;=\qquad\;\frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \hdots + \frac{n-1}{3^{n-1}} + \frac{n}{3^n}

    \displaystyle \text{Subtract: }\qquad\;\; \frac{2}{3}S \;=\;\underbrace{1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \hdots + \frac{1}{3^{n-1}}}_{\text{geometric series}} - \frac{n}{3^n} .[1]


    The geometric series has: first term a = 1, common ratio r = \tfrac{1}{3}, and \,n terms.

    . . Its sum is: . \dfrac{1-(\frac{1}{3})^n}{1-\frac{1}{3}} \;=\;\frac{3}{2}\left(1 - \frac{1}{3^n}\right)


    Then [1] becomes:

    . . \displaystyle \frac{2}{3}S \;=\;\frac{3}{2}\left(1-\frac{1}{3^n}\right) - \frac{n}{3^n} \;=\;\frac{3}{2}\left(3 - \frac{2n+3}{3^n}\right)


    \displaystyle \text{Therefore: }\;S \;=\;\frac{3}{4}\left(3 - \frac{2n+3}{3^n}\right)
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