Question

the sum of infinity of series 1+2/3+6/9+​

Answer 1

Step-by-step explanation:

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Answer 2

Question :

Find the sum to infinity of the series $1\;+\;\frac{2}{3}\;+\;\frac{6}{3^2}\;+\;\frac{10}{3^3}\;+\;\frac{14}{3^4}\;+\;...\;is$

Solution :

Let S = $1\;+\;\frac{2}{3}\;+\;\frac{6}{3^2}\;+\;\frac{10}{3^3}\;+\;\frac{14}{3^4}\;+\;...\rightarrow(1)$

$\frac{1}{3}S\;=\;\frac{1}{3}\;+\;\frac{2}{3^2}\;+\;\frac{6}{3^3}\;+\;\frac{10}{3^4}\;+\;...\rightarrow(2)$

Subtracting (2) from (1),

$S(1\;-\;\frac{1}{3})\;=\;1\;+\;\frac{1}{3}\;+\;\frac{4}{3^2}\;+\frac{4}{3^3}\;+\;\frac{4}{3^4}\;+\;...$

$S(\frac{2}{3})\;=\;\frac{4}{3}\;+\;\frac{4}{3^2}\;(1\;+\;\frac{1}{3}\;+\;\frac{1}{3^2}\;+\;...)$

$\frac{2}{3}S\;=\;\frac{4}{3}\;+\;\frac{4}{3^2}(\frac{1}{1-\frac{1}{3}) }$

$\frac{2}{3}S\;=\;\frac{4}{3}\;+\;\frac{4}{3^2}.\frac{3}{2}\;=\;2$

∴ $S\;=\;3$