Q. If $0 < y < 2^{\frac{1}{3}}$ and $x\left( y^{3} - 1 \right) = 1$, then $\frac{2}{x} + \frac{2}{3x^{3}} + \frac{2}{5x^{5}} + \ldots$ is equal to
A) $\log\left( \frac{y^{3}}{2 - y^{3}} \right)$
B) $\log\left( \frac{y^{3}}{1 - y^{3}} \right)$
C) $\log\left( \frac{2y^{3}}{1 - y^{3}} \right)$
D) $\log\left( \frac{y^{3}}{1 - 2y^{3}} \right)$
Explanation
(a)
$x\left( y^{3} - 1 \right) = 1 \Rightarrow x = \frac{1}{y^{3} - 1} = \frac{1}{k}$ [say]
$\Rightarrow k = \frac{1}{x}$
Then, $\frac{2}{x} + \frac{2}{3x^{3}} + \frac{2}{5x^{5}} + ... = 2k + \frac{2}{3}k^{3} + \frac{2}{5}k^{5} + ...$
$= \log\frac{1 + k}{1 - k} = \log\left\{ \frac{1 + y^{3} - 1}{1 - y^{3} + 1} \right\}$
$= \log\frac{y^{3}}{2 - y^{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.