Question

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

• A. 3
• B. 4
• C. 6
• D. 2

Answer 1

Answer: (A)

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