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} $
$\Rightarrow S=3$