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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
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$\frac{1}{3} y-\frac{1.4}{3^{2} 2} y^{2}+\frac{1.4 .7}{3^{2} 3} y^{3}-\ldots \ldots \infty$ is equal to
Solution:
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Verified Answer
The correct answer is:
$x$
We have $y=3 x+6 x^{2}+10 x^{3}+\ldots \ldots$
$\Rightarrow 1+y=1+3 x+6 x^{2}+10 x^{3}+\ldots$
$\Rightarrow 1+y=(1-x)^{-3} \Rightarrow 1-x=(1+y)^{-1 / 3}$
$\Rightarrow x=1-(1+y)^{-1 / 3}$
$=\frac{1}{3} y-\frac{1.4}{3^{2} \cdot 2} y^{2}+\frac{1 \cdot 4 \cdot 7}{3^{2} \cdot 3} y^{3}-\ldots$
$\Rightarrow 1+y=1+3 x+6 x^{2}+10 x^{3}+\ldots$
$\Rightarrow 1+y=(1-x)^{-3} \Rightarrow 1-x=(1+y)^{-1 / 3}$
$\Rightarrow x=1-(1+y)^{-1 / 3}$
$=\frac{1}{3} y-\frac{1.4}{3^{2} \cdot 2} y^{2}+\frac{1 \cdot 4 \cdot 7}{3^{2} \cdot 3} y^{3}-\ldots$
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