The sum of the infinity of the series 1+23+632+1033+1434+⋯ is

1+23+632+1033+1434+⋯ =1+13[2+(2+4)13+(2+8)132+⋯]=1+13[21−13+4.13(1−13)2] =1+13[3+4349]=1+13(6)=3

The sum of the infinity of the series 1+23+632+1033+1434+⋯ is

1+23+632+1033+1434+⋯ =1+13[2+(2+4)13+(2+8)132+⋯]=1+13[21−13+4.13(1−13)2] =1+13[3+4349]=1+13(6)=3

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The sum of the infinity 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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Detailed Solution

1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \frac{14}{3^{4}} + \dots

$= 1 + \frac{1}{3} [2 + (2 + 4) \frac{1}{3} + (2 + 8) \frac{1}{3^{2}} + \dots] = 1 + \frac{1}{3} [\frac{2}{1 - \frac{1}{3}} + \frac{4. \frac{1}{3}}{{(1 - \frac{1}{3})}^{2}}]$

$= 1 + \frac{1}{3} [3 + \frac{\frac{4}{3}}{\frac{4}{9}}] = 1 + \frac{1}{3} (6) = 3$

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The sum of the infinity of the series 1+23+632+1033+1434+⋯ is

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