Question

$\frac{2}{1}\times \frac{1}{3}+\frac{3}{2}\times \frac{1}{9}+\frac{4}{3}\times \frac{1}{27}+\frac{5}{4}\times \frac{1}{81}+...........$is equal to

• A. $\frac{1}{2}+{\log }_e\left(\frac{3}{2}\right)$
• B. ${\log }_e\left(3\right)-{\log }_e\left(2\right)$
• C. ${\log }_e\left(6\right)$
• D. ${\log }_e\left(2\right)-{\log }_e\left(3\right)$

Answer 1

The correct option is A

$\frac{1}{2}+{\log }_e\left(\frac{3}{2}\right)$

Explanation for correct option:

Step-1: Finding $n^{th}$term of the given series.

Let the given series, $\mathrm{S}=\frac{2}{1}\times \frac{1}{3}+\frac{3}{2}\times \frac{1}{9}+\frac{4}{3}\times \frac{1}{27}+\frac{5}{4}\times \frac{1}{81}+...........$

We can notice $n^{th}$ term of series can be $a_n=\frac{n+1}{n}\times \frac{1}{3^n}$

Step-2: Use sum of infinite G.P. formula

$\begin{array}{rcl}{\mathrm{S}}_{\infty }& =& \sum _{\mathrm{n}=1}^{\infty }{\mathrm{a}}_{\mathrm{n}}\\ & \Rightarrow & {\mathrm{S}}_{\infty }=\sum _{n\mathit{=}\mathit{1}}^{\infty }\frac{n\mathit{+}\mathit{1}}{n}\times \frac{1}{3^n}\\ & \Rightarrow & {\mathrm{S}}_{\infty }=\sum _{n\mathit{=}\mathit{1}}^{\infty }\mathit{(}\mathit{1}\mathit{+}\frac{\mathit{1}}{n}\mathit{)}\times \frac{1}{3^{\mathrm{n}}}\\ & \Rightarrow & {\mathrm{S}}_{\infty }=\sum _{n\mathit{=}\mathit{1}}^{\infty }\frac{1}{3^{\mathrm{n}}}+\sum _{\mathrm{n}\mathit{=}\mathit{1}}^{\infty }\frac{1}{n{3}^{\mathrm{n}}}\\ & \Rightarrow & {\mathrm{S}}_{\infty }=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....\infty \right)+\left(\frac{1}{1\times 3^1}+\frac{1}{2\times 3^2}+\frac{1}{3\times 3^3}+....\infty \right)\\ & \Rightarrow & {\mathrm{S}}_{\infty }=\frac{{\displaystyle \frac{1}{3}}}{\left(1-{\displaystyle \frac{1}{3}}\right)}+\left(\frac{1}{1\times 3^1}+\frac{1}{2\times 3^2}+\frac{1}{3\times 3^3}+....\infty \right)\left(\because SumofinfiniteG.P.formula{\mathrm{S}}_{\infty }=\frac{\mathrm{a}}{1-\mathrm{r}}\right)\end{array}$

Step 3: Use log expansion formula

We know that,

${\log }_e\left(1-x\right)=-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+...........\right)$

$\begin{array}{rcl}& \Rightarrow & {\log }_{\mathrm{e}}\left(1-\frac{1}{3}\right)=\frac{1}{1\times 3^1}+\frac{1}{2\times 3^2}+\frac{1}{3\times 3^3}+....\infty \end{array}$

$\Rightarrow {\mathrm{S}}_{\infty }=\frac{{\displaystyle \frac{1}{3}}}{\left({\displaystyle \frac{3-1}{3}}\right)}+\left(-{\log }_{\mathrm{e}}\left(1-\frac{1}{3}\right)\right)$

Therefore,

${\mathrm{S}}_{\infty }=\frac{1}{2}+{\log }_{\mathrm{e}}\left(\frac{3}{2}\right)$

Thus, correct answer is option $\left(A\right)$.