Question

$\frac{1}{2}{x}^2+\frac{2}{3}{x}^3+\frac{3}{4}{x}^4+\frac{4}{5}{x}^5+\dots \dots =$

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

Answer 1

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

Let $y=\frac{1}{2}{x}^2+\frac{2}{3}{x}^3+\frac{3}{4}{x}^4+\frac{4}{5}{x}^5+\dots \dots$
$\Rightarrow y=(1-\frac{1}{2})x^2+(1-\frac{1}{3})x^3+(1-\frac{1}{4})x^4+(1-\frac{1}{5})x^5+\dots$
$\Rightarrow y=(x^2+x^3+x^4+\dots )-(\frac{1}{2}{x}^2+\frac{1}{3}{x}^3+\frac{1}{4}{x}^4+\frac{1}{5}{x}^5+\dots )$
$\Rightarrow y=\frac{x^2}{1-x}-\left(\log \right(1-x)-x)=\frac{x}{1-x}+\log (1-x)$
Note: Here assumption is that $|x|<1$