Summation of Arithmetico-Geometrical sequence........

**Arka** · October 22nd 2010, 06:53 PM

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 .

**Soroban** · October 22nd 2010, 09:03 PM

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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