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)

$3$

Answer 2

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