Question

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

• A. $3$
• B. $4$
• C. $6$
• D. $2$

Answer 1

The correct option is A

$3$

Explanation for the correct option:

Find the required sum of given series.

Given:$1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+...$

Let $S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+...$ …(i)
Multiplying by $\frac{1}{3}$​, we get
$\frac{1}{3}S=\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+...$ …(ii)

(i)−(ii), gives

$$\begin{gathered} S–\frac{S}{3}=(1+\left(\frac{2}{3}–\frac{1}{3}\right)+\left(\frac{6}{3^2}–\frac{2}{3^2}\right)+\left(\frac{10}{3^3}–\frac{6}{3^3}\right)+\dots \\ \Rightarrow \frac{2}{3}S=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+... \\ \Rightarrow \frac{2}{3}S=\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+... \end{gathered}$$

$\Rightarrow \frac{2}{3}S=\frac{4}{3}\left(\frac{1}{1-{\displaystyle \frac{1}{3}}}\right)$ [ Sum of infinity GP $=a\left(\frac{1}{1-r}\right)$ where $r<1$]

$$\begin{gathered} \Rightarrow \frac{2}{3}S=\frac{4}{3}\times \frac{3}{2} \\ \Rightarrow \frac{2}{3}S=2 \\ \Rightarrow S=3 \end{gathered}$$

Hence, option (A) is the correct answer.