Q. If $y = 3x + 6x^{2} + 10x^{3} + ...$, then the value of $x$ in terms of $y$ is
A) $1 - (1 - y)^{\frac{- 1}{3}}$
B) $1 - (1 + y)^{\frac{1}{3}}$
C) $1 + (1 + y)^{\frac{- 1}{3}}$
D) $1 - (1 + y)^{\frac{- 1}{3}}$
Explanation
(d)
Given, $y = 3x + 6x^{2} + 10x^{3} + ...$
$\therefore 1 + y = 1 + 3x + 6x^{2} + 10x^{3} + ...$
$\Rightarrow 1 + y = (1 - x)^{- 3}$
$\Rightarrow 1 - x = (1 + y)^{- 1/3}$
$\Rightarrow x = 1 - (1 + y)^{- 1/3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.