Question

The sum to infinity of the series $1+\left(1+\frac{1}{2}\right)\left(\frac{1}{3}\right)+\left(1+\frac{1}{2}+\frac{1}{2^2}\right)\left(\frac{1}{3^2}\right)+\dots \dots \infty$ is

• A. $\frac{12}{5}$
• B. $\frac{9}{5}$
• C. $\frac{9}{10}$
• D. 3

Answer 1

Answer: (B)

$t_n=\left(1+\frac{1}{2}+\frac{1}{2^2}+\dots ...+\frac{1}{2^{n-1}}\right)\frac{1}{3^{n-1}}=\left(\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}\right)\frac{1}{3^{n-1}}=\frac{2^n-1}{6^{n-1}}=2{\left(\frac{1}{3}\right)}^{n-1}-{\left(\frac{1}{6}\right)}^{n-1}$
$S_{\infty }=\frac{2}{1-\frac{1}{3}}-\frac{1}{1-\frac{1}{6}}=3-\frac{6}{5}=\frac{9}{5}$