Question

The sum of the infinite series $1+\frac{2}{3}+\frac{7}{3^2}+\frac{12}{3^3}+\frac{17}{3^4}+\frac{22}{3^5}+\dots$. is equal to

• A. $\frac{9}{4}$
• B. $\frac{15}{6}$
• C. $\frac{13}{6}$
• D. $\frac{11}{6}$

Answer 1

The correct option is C

$\frac{13}{6}$

The explanation for the correct option:

Given, $1+\frac{2}{3}+\frac{7}{3^2}+\frac{12}{3^3}+\frac{17}{3^4}+\frac{22}{3^5}+\dots$

Let sum be $S$

$$\begin{gathered} \Rightarrow S=1+\frac{2}{3}+\frac{7}{3^2}+\frac{12}{3^3}+\frac{17}{3^4}+\frac{22}{3^5}+\dots ..\infty ...\left(1\right) \\ \Rightarrow \frac{S}{3}=\frac{1}{3}+\frac{2}{3^2}+\frac{7}{3^3}+\frac{12}{3^4}+..........\infty ...\left(2\right) \end{gathered}$$

Subtracting $\left(2\right)$ from $\left(1\right)$

$$\begin{gathered} \Rightarrow \frac{2S}{3}=1+\frac{1}{3}+\frac{5}{3^2}+\frac{5}{3^3}+.........\infty \\ \Rightarrow \frac{2S}{3}=\frac{4}{3}+\left(\frac{5}{3^2}+\frac{5}{3^3}+.........\infty \right) \end{gathered}$$

Here, the first term of the G.P.$=\frac{5}{3^2}$

The common ratio$=\frac{1}{3}$

$$\begin{gathered} \Rightarrow \frac{2S}{3}=\frac{4}{3}+\left(\frac{{\displaystyle \frac{5}{3^2}}}{1-{\displaystyle \frac{1}{3}}}\right)\left[\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{infinite}\mathrm{GP}=\frac{a}{1-r}\right] \\ \Rightarrow S=\frac{4}{3}+\frac{5}{6} \\ =\frac{13}{6} \end{gathered}$$

Hence, the correct option is (C).