The sum of series 1+45+752+1053+⋯∞ is

LetS=1+45+752+1053+⋯∞Then15S=15+452+753+⋯∞or S1−15=1+315+152+153+⋯∞or 45S=1+3×1/51−(1/5)=1+34=74or S=3516

The sum of series 1+45+752+1053+⋯∞ is

LetS=1+45+752+1053+⋯∞Then15S=15+452+753+⋯∞or S1−15=1+315+152+153+⋯∞or 45S=1+3×1/51−(1/5)=1+34=74or S=3516

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The sum of series $1 + \frac{4}{5} + \frac{7}{5^{2}} + \frac{10}{5^{3}} + \dots \infty$ is

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7/16

5/16

105/64

35/16

answer is D.

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Detailed Solution

Let

$S = 1 + \frac{4}{5} + \frac{7}{5^{2}} + \frac{10}{5^{3}} + \dots \infty$

Then

$\frac{1}{5} S = \frac{1}{5} + \frac{4}{5^{2}} + \frac{7}{5^{3}} + \dots \infty$

$\begin{array}{l} or S (1 - \frac{1}{5}) = 1 + 3 [\frac{1}{5} + \frac{1}{5^{2}} + \frac{1}{5^{3}} + \dots \infty] \\ or \frac{4}{5} S = 1 + 3 \times \frac{1 / 5}{1 - (1 / 5)} = 1 + \frac{3}{4} = \frac{7}{4} \\ or S = \frac{35}{16} \end{array}$

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The sum of series 1+45+752+1053+⋯∞ is

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The sum of series $1 + \frac{4}{5} + \frac{7}{5^{2}} + \frac{10}{5^{3}} + \dots \infty$ is