Question

The sum of the infinite series $1+(1+\frac{1}{2})\left(\frac{1}{3}\right)+(1+\frac{1}{2}+\frac{1}{2^2})\left(\frac{1}{3^2}\right)+.....\infty$

• A. $\frac{12}{5}$
• B. $\frac{9}{5}$
• C. $\frac{8}{5}$
• D. $\frac{5}{3}$

Answer 1

The correct option is B $\frac{9}{5}$

Let, $S_{\infty }=1+(1+\frac{1}{2})\left(\frac{1}{3}\right)+(1+\frac{1}{2}+\frac{1}{2^2})\left(\frac{1}{3^2}\right)+.....\infty$
Now, $t_r=(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{r-1}})\left(\frac{1}{3^{r-1}}\right)=\left(\frac{1-{\left(1/2\right)}^r}{1-\left(1/2\right)}\right)\left(\frac{1}{3^{r-1}}\right)$ ...[ Sum of G.P series ]
$\Rightarrow t_r=6(\frac{1}{3^r}-\frac{1}{6^r})$
$S_{\infty }=\sum _{r=1}^{\infty }{t}_r=6\sum _{r=1}^{\infty }(\frac{1}{3^r}-\frac{1}{6^r})$
$\Rightarrow S_{\infty }=6\left([\frac{1/3}{1-\left(1/3\right)}-\frac{1/6}{1-\left(1/6\right)}]\right)$ .....[ Sum of Infinite G.P series ]
$\Rightarrow S_{\infty }=\frac{9}{5}$

Ans: B