Question

The value of $\sum _{n=1}^{\infty }\left(\frac{4n-2}{3^n}\right)$ is equal to

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

$S=\sum _{n=1}^{\infty }(\frac{4n}{3^n}-\frac{2}{3^n})=$$4\sum _{n=1}^{\infty }\left(\frac{n}{3^n}\right)$$-2\sum _{n=1}^{\infty }\left(\frac{1}{3^n}\right)$

Formula for sum of AGP for infinite series

$\Rightarrow \frac{a}{1-r}+\frac{dr}{(1-r{)}^2}$

$a=\frac{1}{3},r=\frac{1}{3},d=1$

$\Rightarrow \sum _{n=1}^{\infty }\frac{n}{3^n}=\frac{1/3}{2/3}+\frac{1/3}{4/9}=\frac{1}{3}\times \frac{3}{2}+\frac{1}{3}\times \frac{9}{4}=\frac{1}{2}+\frac{3}{4}=\frac{5}{4}$

$\Rightarrow \sum _{n=1}^{\infty }\frac{1}{3^n}=\frac{a}{1-r}=\frac{1/3}{2/3}=\frac{1}{3}\times \frac{3}{2}=\frac{1}{2}$

$\Rightarrow S=4\times \frac{5}{4}-2\times \frac{1}{2}=4$