Question

$\frac{1}{2^1}-\frac{1}{2.2^2}+\frac{1}{3.2^3}-\frac{1}{4.2^4}+....isequalto$

• A. $\frac{1}{4}$
• B. ${\log }_e\left(\frac{3}{4}\right)$
• C. ${\log }_e\left(\frac{3}{2}\right)$
• D. ${\log }_e\left(\frac{2}{3}\right)$

Answer 1

The correct option is C

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

Explanation for the correct option:

Find the value of given series using expansion formula of log

As we know,

${\mathbf{log}}_{\mathbf{e}}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathit{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{–}\mathbf{}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{4}}\mathbf{+}\mathbf{\dots }$

Put $x=\frac{1}{2}$,

$\Rightarrow {\log }_e\left(1+\frac{1}{2}\right)=\frac{1}{2}-\frac{{\left({\displaystyle \frac{1}{2}}\right)}^2}{2}+\frac{{\left({\displaystyle \frac{1}{2}}\right)}^3}{3}–\frac{{\left({\displaystyle \frac{1}{2}}\right)}^4}{4}+\dots$

$\Rightarrow {\log }_e\left(\frac{3}{2}\right)=\frac{1}{2}-\frac{{\left({\displaystyle \frac{1}{2}}\right)}^2}{2}+\frac{{\left({\displaystyle \frac{1}{2}}\right)}^3}{3}–\frac{{\left({\displaystyle \frac{1}{2}}\right)}^4}{4}+\dots$

$\Rightarrow \frac{1}{2^1}-\frac{1}{2.2^2}+\frac{1}{3.2^3}-\frac{1}{4.2^4}+....={\mathbf{log}}_{\mathbf{e}}\mathbf{\left(}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{\right)}$

Hence, Option ‘C’ is Correct.