Question

If $0<x<1$, then $\frac{3}{2}{x}^2+\frac{5}{3}{x}^3+\frac{7}{4}{x}^4+...,$ is equal to

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

Answer 1

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

Given: $x\in (0,1)$
Let $E=\frac{3}{2}{x}^2+\frac{5}{3}{x}^3+\frac{7}{4}{x}^4+\dots \infty$
$\Rightarrow E=(2-\frac{1}{2})x^2+(2-\frac{1}{3})x^3+(2-\frac{1}{4})x^4+\dots \infty$
$\Rightarrow E=2x^2(1+x+x^2+\dots \infty )-(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\dots \infty )$
$\Rightarrow E=2x^2\cdot \frac{1}{1-x}+x-(\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\dots \infty )$
$\therefore E=x\left(\frac{1+x}{1-x}\right)+\ln (1-x)$