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

The correct option is D $1-(1+y{)}^{-1/3}$

Let $f\left(x\right)=(1-x{)}^{-3}$

$\Rightarrow f^{\prime }\left(x\right)=3(1-x{)}^{-4}$

$\Rightarrow f^{\prime \prime }\left(x\right)=3\times 4(1-x{)}^{-5}$

$\Rightarrow f^{\prime \prime \prime }\left(x\right)=3\times 4\times 5(1-x{)}^{-6}$

So, $f\left(0\right)=1,f^{\prime }\left(0\right)=3=\frac{3!}{2!},f^{\prime \prime }\left(0\right)=3\times 4=\frac{4!}{2!},....$

$f\left(x\right)=\sum _{i=0}^{\infty }\frac{(2+i)!}{2!}\times \frac{x^n}{n!}$

$\therefore f\left(x\right)=1+3x+6x^2+10x^3+...$

Now,

$\because$ $y=3x+6x^2+10x^3+...$

$\Rightarrow$ $1+y=1+3x+6x^2+10x^3+...$

$\Rightarrow 1+y=(1-x{)}^{-3}$

$\Rightarrow 1-x=(1+y{)}^{-1/3}$

$\Rightarrow x=1-(1+y{)}^{-1/3}$