Question

The sum to infinity of the series $2+\frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^3}+\dots$ will be

• A. $3$
• B. $4$
• C. $\frac{7}{2}$
• D. $\frac{9}{2}$

Answer 1

The correct option is C

$\frac{7}{2}$

Explanation for the correct answer:

Let, $S=2+\frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^3}+\dots$

Group the fractions containing the same base

$\Rightarrow S=2+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots \right)+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\dots \right)$

We see that the sum has two geometric series in it. The formula to find the sum of a geometric series is,

$S=\frac{a}{1-r}$

where, $a$ is the first term and $r$ is the common ratio.

In the above sum, the first term and common ratio of the first series is $\frac{1}{2}$, and the first term and common ratio of the second series is $\frac{1}{3}$. Thus,

$$\begin{gathered} \Rightarrow S=2+\frac{{\displaystyle \frac{1}{2}}}{1-{\displaystyle \frac{1}{2}}}+\frac{{\displaystyle \frac{1}{3}}}{1-{\displaystyle \frac{1}{3}}} \\ \Rightarrow S=2+\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}}+\frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{2}{3}}} \\ \Rightarrow S=2+1+\frac{1}{2} \\ \Rightarrow S=3\frac{1}{2}=\frac{7}{2} \end{gathered}$$

Hence, $\mathrm{Option}\left(\mathrm{C}\right)$ is correct.