What is the sum of the series 1 + frac{2}{3} | vedclass

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(D) Let $S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots \infty \dots (1)$Multiplying by $\frac{1}{3}$,we get $\frac{1}{

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$3$

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(D) Let $S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots \infty \dots (1)$Multiplying by $\frac{1}{3}$,we get $\frac{1}{3}S = \frac{1}{3} + \frac{2}{3^2} + \frac{6}{3^3} + \frac{10}{3^4} + \dots \infty \dots (2)$Subtracting $(2)$ from $(1)$:$S - \frac{1}{3}S = 1 + (\frac{2}{3} - \frac{1}{3}) + (\frac{6}{3^2} - \frac{2}{3^2}) + (\frac{10}{3^3} - \frac{6}{3^3}) + \dots$$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{4}{3^2} + \frac{4}{3^3} + \frac{4}{3^4} + \dots$$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{4}{3^2} (1 + \frac{1}{3} + \frac{1}{3^2} + \dots)$Using the sum of an infinite geometric series formula $S_{\infty} = \frac{a}{1-r}$:$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{4}{9} (\frac{1}{1 - 1/3}) = 1 + \frac{1}{3} + \frac{4}{9} (\frac{3}{2}) = 1 + \frac{1}{3} + \frac{2}{3} = 2$$\frac{2}{3}S = 2 \implies S = 3$.

What is the sum of the series $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$?

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