The sum to infinity of the series 1+23+632+10333+1434+…..is

Solution:

Let S be the sum of the given series i.e.,

$\begin{aligned} S = 1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \frac{14}{3^{4}} + \dots . to \infty \\ \Rightarrow & (S - 1) = \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \frac{14}{3^{4}} + \dots to \infty - - - - (i) \\ \Rightarrow & (S - 1) \times \frac{1}{3} = \frac{2}{3^{2}} + \frac{6}{3^{3}} + \frac{10}{3^{4}} + \dots to \infty - - - - - (ii) \end{aligned}$

Subtracting (ii) from (i), we get

$\frac{2}{3} (S - 1) = \frac{2}{3} + \frac{4}{3^{2}} + \frac{4}{3^{3}} + \frac{4}{3^{4}} + \dots to \infty$

$\Rightarrow \frac{2}{3} (S - 1) = \frac{2}{3} + \frac{\frac{4}{3^{2}}}{1 - \frac{1}{3}} \Rightarrow \frac{2}{3} (S - 1) = \frac{2}{3} + \frac{2}{3} \Rightarrow S = 3$

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

Solution:

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