If S = frac{7}{5} + frac{9}{5^{2}} + frac{13} | vedclass

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(B) Given $S = \frac{7}{5} + \frac{9}{5^{2}} + \frac{13}{5^{3}} + \frac{19}{5^{4}} + \ldots$ $(1)$Multiply by $\frac{1}{5}$:$\frac{1}{5} S = \frac{7}{

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$305$

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(B) Given $S = \frac{7}{5} + \frac{9}{5^{2}} + \frac{13}{5^{3}} + \frac{19}{5^{4}} + \ldots$ $(1)$Multiply by $\frac{1}{5}$:$\frac{1}{5} S = \frac{7}{5^{2}} + \frac{9}{5^{3}} + \frac{13}{5^{4}} + \ldots$ $(2)$Subtracting $(2)$ from $(1)$:$S - \frac{1}{5} S = \frac{7}{5} + \left(\frac{9-7}{5^{2}}ight) + \left(\frac{13-9}{5^{3}}ight) + \left(\frac{19-13}{5^{4}}ight) + \ldots$$\frac{4}{5} S = \frac{7}{5} + \frac{2}{5^{2}} + \frac{4}{5^{3}} + \frac{6}{5^{4}} + \ldots$Let $T = \frac{2}{5^{2}} + \frac{4}{5^{3}} + \frac{6}{5^{4}} + \ldots$Then $\frac{1}{5} T = \frac{2}{5^{3}} + \frac{4}{5^{4}} + \ldots$Subtracting: $\frac{4}{5} T = \frac{2}{5^{2}} + \frac{2}{5^{3}} + \frac{2}{5^{4}} + \ldots = \frac{2/25}{1 - 1/5} = \frac{2/25}{4/5} = \frac{1}{10}$So $T = \frac{5}{4} 	imes \frac{1}{10} = \frac{1}{8}$$\frac{4}{5} S = \frac{7}{5} + \frac{1}{8} = \frac{56 + 5}{40} = \frac{61}{40}$$S = \frac{61}{40} 	imes \frac{5}{4} = \frac{61}{32}$$160 \,S = 160 	imes \frac{61}{32} = 5 	imes 61 = 305$

If $S = \frac{7}{5} + \frac{9}{5^{2}} + \frac{13}{5^{3}} + \frac{19}{5^{4}} + \ldots$, then $160 \,S$ is equal to....... .

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