0 votes

The value of $\dfrac{x}{1-x^2} + \dfrac{x^2}{1-x^4} + \dfrac{x^4}{1-x^8} + \dfrac{x^8}{1-x^{16}}$ is

- $\frac{1}{1-x^{16}}$
- $\frac{1}{1-x^{12}}$
- $\frac{1}{1-x} β \frac{1}{1-x^{16}}$
- $\frac{1}{1-x} β \frac{1}{1-x^{12}}$

1 vote

Adding and Subtracting 1 on each term

$S= \frac{1+x -1}{1-x^{2}}+\frac{1+x^{2} -1}{1-x^{4}}+\frac{1+x^{4} -1}{1-x^{8}}+\frac{1+x^{8} -1}{1-x^{16}}$

Using $(a^{2}-b^{2})=(a+b)(a-b)$

$S=\frac{1}{1-x}-\frac{1}{1-x^{2}}+\frac{1}{1-x^{2}}-\frac{1}{1-x^{4}}+\frac{1}{1-x^{4}}-\frac{1}{1-x^{8}}+\frac{1}{1-x^{8}}-\frac{1}{1-x^{16}}$

Simplifies to $S = \frac{1}{1-x} - \frac{1}{1-x^{16}}$

**Answer: C**