Question

$\frac{1}{2}(\frac{1}{2}{)}^2+\frac{1}{4}(\frac{1}{2}{)}^4+\frac{1}{6}(\frac{1}{2}{)}^6+\dots \dots =$

• A. $lo{g}_e\frac{4}{3}$
• B. $-\frac{1}{2}lo{g}_e\frac{4}{3}$
• C. $lo{g}_e\frac{3}{4}$
• D. $-\frac{1}{2}lo{g}_e\frac{3}{4}$

Answer 1

The correct option is D $-\frac{1}{2}lo{g}_e\frac{3}{4}$

Let S=$\frac{1}{2}(\frac{1}{2}{)}^2+\frac{1}{4}(\frac{1}{2}{)}^4+\frac{1}{6}(\frac{1}{2}{)}^6+\dots \dots$
Put $X=\frac{1}{2}$
$S=\frac{1}{2}[X^2+\frac{1}{2}{X}^4+\frac{1}{3}{x}^6+.....]$
$S=\frac{1}{2}[X^2+\frac{(X^2{)}^2}{2}+\frac{(X^2{)}^3}{3}+.....]$

$S=-\frac{1}{2}[-X^2-\frac{(X^2{)}^2}{2}-\frac{(X^2{)}^3}{3}-.....]$
We know $-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}......={\log }_e(1-x)=-\sum \frac{x^n}{n}$
$\Rightarrow S=-\frac{1}{2}{\log }_e(1-X^2)$
$\Rightarrow S=-\frac{1}{2}{\log }_e\left(\frac{3}{4}\right)$