Question

The sum of the series, $\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+....$ to infinite terms if $|x|<1$ is

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

Answer 1

The correct option is A $\frac{x}{1-x}$

$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+.......\infty \frac{x}{1-x^2}=\frac{1}{1-x}-\frac{1}{1-x^2}\frac{x^2}{1-x^4}=\frac{1}{1-x^2}-\frac{1}{1-x^4}\frac{x^4}{1-x^6}=\frac{1}{1-x^4}-\frac{1}{1-x^8}.......\infty$

AS $\left|x\right|<1$ so last term would tend to $1$.

So, all terms gets cancelled out except,$\frac{1}{1-x}-1$

$=\frac{1-1+x}{1-x}=\frac{x}{1-x}$

Answer $\left(A\right)$