The sum of infinite terms of the following se | vedclass

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Question

(D) Let the sum to infinity of the arithmetico-geometric series be $S = 1 + 4 \cdot \frac{1}{5} + 7 \cdot \frac{1}{5^2} + 10 \cdot \frac{1}{5^3} + \do

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$\frac{35}{16}$

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(D) Let the sum to infinity of the arithmetico-geometric series be $S = 1 + 4 \cdot \frac{1}{5} + 7 \cdot \frac{1}{5^2} + 10 \cdot \frac{1}{5^3} + \dots$$\frac{1}{5}S = \frac{1}{5} + 4 \cdot \frac{1}{5^2} + 7 \cdot \frac{1}{5^3} + \dots$Subtracting the two equations:$(1 - \frac{1}{5})S = 1 + 3 \cdot \frac{1}{5} + 3 \cdot \frac{1}{5^2} + 3 \cdot \frac{1}{5^3} + \dots$$\frac{4}{5}S = 1 + 3 \left( \frac{1}{5} + \frac{1}{5^2} + \dots ight)$Using the sum of an infinite geometric series formula $S_{\infty} = \frac{a}{1-r}$ where $a = \frac{1}{5}$ and $r = \frac{1}{5}$:$\frac{4}{5}S = 1 + 3 \left( \frac{1/5}{1 - 1/5} ight) = 1 + 3 \left( \frac{1/5}{4/5} ight) = 1 + 3 \left( \frac{1}{4} ight) = 1 + \frac{3}{4} = \frac{7}{4}$$S = \frac{7}{4} 	imes \frac{5}{4} = \frac{35}{16}$.

The sum of infinite terms of the following series $1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \dots$ will be

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