Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

Answer: (B)

Let $S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{10}{3^3}+\frac{14}{3^4}+...\left(1\right)$
$\frac{1}{3}S=\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+...\left(2\right)$
Dividing $\left(1\right)$
$S\left(1-\frac{1}{3}\right)=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+....$
$\frac{2}{3}S=\frac{4}{3}+\frac{4}{3^2}\left(1+\frac{1}{3}+\frac{1}{3^2}+.....\right)\Rightarrow \frac{2}{3}$$S=\frac{4}{3}+\frac{4}{3^2}\left(\frac{1}{1-\frac{1}{3}}\right)=\frac{4}{3}+\frac{4}{3^2}\frac{3}{2}+\frac{4}{3}+\frac{2}{3}=\frac{6}{2}\Rightarrow \frac{2}{3}S=\frac{6}{3}\Rightarrow S=3$