Q75 / 125 · BITSAT 2020 · Mathematics

If $y=3 x+6 x^{2}+10 x^{3}+\ldots \ldots \ldots, \infty,$ then $\frac{1}{3} y-\frac{1.4}{3^{2} 2} y^{2}+\frac{1.4 .7}{3^{2} 3} y^{3}-\ldots \ldots \infty$ is equal to

  • Correct answer
    A
    $x$
  • B
    $1-x$
  • C
    $1+x$
  • D
    $x^{x}$

Solution

We have $y=3 x+6 x^{2}+10 x^{3}+\ldots \ldots$ $\Rightarrow 1+y=1+3 x+6 x^{2}+10 x^{3}+\ldots$ $\Rightarrow 1+y=(1-x)^{-3} \Rightarrow 1-x=(1+y)^{-1 / 3}$ $\Rightarrow x=1-(1+y)^{-1 / 3}$ $=\frac{1}{3} y-\frac{1.4}{3^{2} \cdot 2} y^{2}+\frac{1 \cdot 4 \cdot 7}{3^{2} \cdot 3} y^{3}-\ldots$

Hence, the correct answer is option: 'A'.

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