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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2020 · Mathematics · Sequences and Series

If y=3 x+6 x^ 2 +10 x^ 3 + , ∞, then (1)/(3) y- 1.4 3^ 2 2 y^ 2 +...

If y=3x+6x2+10x3+………,∞,y=3 x+6 x^{2}+10 x^{3}+\ldots \ldots \ldots, \infty, then 13y−1.4322y2+1.4.7323y3−……∞\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

Options

  1. Axx
  2. B1−x1-x
  3. C1+x1+x
  4. Dxxx^{x}

Correct answer

A. xx

Step-by-step solution

We have y=3x+6x2+10x3+……⇒1+y=1+3x+6x2+10x3+…⇒1+y=(1−x)−3⇒1−x=(1+y)−1/3⇒x=1−(1+y)−1/3=13y−1.432⋅2y2+1⋅4⋅732⋅3y3−…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
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