The sum to infinite terms of the series 1 + f | vedclass

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(A) Let the sum be $S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ $(1)$Multiply by $\frac{1}{3}$: $\frac{1}{3}S = \fr

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$3$

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(A) Let the sum be $S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ $(1)$Multiply by $\frac{1}{3}$: $\frac{1}{3}S = \frac{1}{3} + \frac{2}{3^2} + \frac{6}{3^3} + \frac{10}{3^4} + \dots$ $(2)$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}) + (\frac{14}{3^4} - \frac{10}{3^4}) + \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} + \frac{4}{3^3} + \frac{4}{3^4} + \dots]$The term in the bracket is an infinite geometric series with $a = \frac{4}{9}$ and $r = \frac{1}{3}$.Sum $= \frac{a}{1-r} = \frac{4/9}{1 - 1/3} = \frac{4/9}{2/3} = \frac{4}{9} 	imes \frac{3}{2} = \frac{2}{3}$.So,$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{2}{3} = 1 + 1 = 2$.$S = 2 	imes \frac{3}{2} = 3$.

The sum to infinite terms of the series $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ is

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