Question

If $y=3x+6x^2+10x^3+...$ , then the value of $x$ in terms of $y$ is

• A. $1-(1-y{)}^{-1/3}$
• B. $1-(1+y{)}^{1/3}$
• C. $1+(1+y{)}^{-1/3}$
• D. $1-(1+y{)}^{-1/3}$

Answer 1

Answer: (D)

Given, $y=3x+6x^2+10x^3+\dots$
$\therefore 1+y=1+3x+6x^2+10x^3+\dots$
$\Rightarrow 1+y=(1-x{)}^{-3}$
$\Rightarrow 1-x=(1+y{)}^{-1/3}$
$\Rightarrow x=1-(1+y{)}^{-1/3}$