Question

The sum to infinite terms of the series
$1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}\cdots$ is

• A. 3
• B. 4
• C. 6
• D. 2

Answer 1

The correct option is A 3

We have
$S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\cdots \infty ...\left(i\right)$
Multiplying both sides by $\frac{1}{3}$ we get
$\frac{1}{3}S=\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+\cdots \infty ...\left(ii\right)$
Subtracting eqn. (ii) from eqn. (i) we get
$\frac{2}{3}S=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\cdots \infty \Rightarrow \frac{2}{3}S=\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\cdots \infty \Rightarrow \frac{2}{3}S=\frac{\frac{4}{3}}{1-\frac{1}{3}}=\frac{4}{3}\times \frac{3}{2}\Rightarrow S=3$