The sum of the infinite series 1+frac{2}{3}+f | vedclass

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Question

(A) Let the sum be $S = 1 + \frac{2}{3} + \frac{7}{3^{2}} + \frac{12}{3^{3}} + \frac{17}{3^{4}} + \frac{22}{3^{5}} + \ldots$ Multiply by $\frac{1}{3}$

Vedclass · Accepted Answer

$\frac{13}{4}$

Vedclass · Answer

(A) Let the sum be $S = 1 + \frac{2}{3} + \frac{7}{3^{2}} + \frac{12}{3^{3}} + \frac{17}{3^{4}} + \frac{22}{3^{5}} + \ldots$ Multiply by $\frac{1}{3}$: $\frac{S}{3} = \frac{1}{3} + \frac{2}{3^{2}} + \frac{7}{3^{3}} + \frac{12}{3^{4}} + \frac{17}{3^{5}} + \ldots$ Subtracting the two equations: $S - \frac{S}{3} = 1 + (\frac{2}{3} - \frac{1}{3}) + (\frac{7}{3^{2}} - \frac{2}{3^{2}}) + (\frac{12}{3^{3}} - \frac{7}{3^{3}}) + \ldots$ $\frac{2S}{3} = 1 + \frac{1}{3} + \frac{5}{3^{2}} + \frac{5}{3^{3}} + \frac{5}{3^{4}} + \ldots$ $\frac{2S}{3} = 1 + \frac{1}{3} + \frac{\frac{5}{3^{2}}}{1 - \frac{1}{3}} = 1 + \frac{1}{3} + \frac{5}{9} 	imes \frac{3}{2} = \frac{4}{3} + \frac{5}{6} = \frac{8+5}{6} = \frac{13}{6}$ $S = \frac{13}{6} 	imes \frac{3}{2} = \frac{13}{4}$

The sum of the infinite series $1+\frac{2}{3}+\frac{7}{3^{2}}+\frac{12}{3^{3}}+\frac{17}{3^{4}}+\frac{22}{3^{5}}+\ldots$ is equal to

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