Question

$x+\frac{5x^3}{2.3}+\frac{9x^5}{4.5}+\frac{13x^7}{6.7}+...to\infty =$

• A. ${\log }_e\left[\frac{(1+x{)}^{\frac{1-x}{2}}}{(1-x{)}^{\frac{1+x}{2}}}\right]$
• B. ${\log }_e\left[\frac{(1-x{)}^{\frac{1+x}{2}}}{(1+x{)}^{\frac{1-x}{2}}}\right]$
• C. ${\log }_e\left[\frac{(1+x{)}^{\frac{1-x}{4}}}{(1-x{)}^{\frac{1+x}{4}}}\right]$
• D. ${\log }_e\left[\frac{(1-x{)}^{\frac{1+x}{4}}}{(1+x{)}^{\frac{1-x}{4}}}\right]$

Answer 1

The correct option is A ${\log }_e\left[\frac{(1+x{)}^{\frac{1-x}{2}}}{(1-x{)}^{\frac{1+x}{2}}}\right]$

Consider ${\log }_e\left[\frac{(1+x{)}^{\frac{1-x}{2}}}{(1-x{)}^{\frac{1+x}{2}}}\right]$
$=\frac{(1-x)}{2}{\log }_e(1+x)-\frac{(1+x)}{2}{\log }_e(1-x)$
$=\frac{1}{2}\left[{\log }_e\right(1+x)-{\log }_e(1-x\left)\right]-\frac{x}{2}\left[{\log }_e\right(1+x)+{\log }_e(1-x\left)\right]$
$=(x+\frac{x^3}{3}+\frac{x^5}{5}+.....)+x(\frac{x^2}{2}+\frac{x^4}{4}+.....)$
$=(x+\frac{x^3}{3}+\frac{x^5}{5}+.....)+(\frac{x^3}{2}+\frac{x^5}{4}+.....)$
$=x+\frac{5x^3}{2.3}+\frac{9x^5}{4.5}+\frac{13x^7}{6.7}+...to\infty$