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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

  1. $\frac{1}{1-x^{16}}$
  2. $\frac{1}{1-x^{12}}$
  3. $\frac{1}{1-x} – \frac{1}{1-x^{16}}$
  4. $\frac{1}{1-x} – \frac{1}{1-x^{12}}$
in Quantitative Aptitude
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1 Answer

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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

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