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The sum to infinite term of the series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\cdots$ is
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The correct answer is:
3
3
We have
Multiplying both sides by $\frac{1}{3}$, we get
Subtracting eqn. (ii) from eqn. (i), we get
$\begin{aligned} & \frac{2}{3} S=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \ldots \ldots \infty \\ & \Rightarrow \frac{2}{3} S=\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \ldots \ldots \infty \\ & \Rightarrow \frac{2}{3} S=\frac{\frac{4}{3}}{1-\frac{1}{3}}=\frac{4}{3} \times \frac{3}{2} \Rightarrow S=3\end{aligned}$
Multiplying both sides by $\frac{1}{3}$, we get
Subtracting eqn. (ii) from eqn. (i), we get
$\begin{aligned} & \frac{2}{3} S=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \ldots \ldots \infty \\ & \Rightarrow \frac{2}{3} S=\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \ldots \ldots \infty \\ & \Rightarrow \frac{2}{3} S=\frac{\frac{4}{3}}{1-\frac{1}{3}}=\frac{4}{3} \times \frac{3}{2} \Rightarrow S=3\end{aligned}$
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