If x2y=2x-y then y2+y42+y63+……∞=

x2y=2x-y ⇒ y(1+x2)=2x ⇒ y=2x1+x2 y2+y222+y233+......∞ =-log (1-y)2 =-log 1-(2x)2(1+x2)2 =-log (1-x2)2(1+x2)2 =-2log 1-x21+x2 =2log 1+x21-x2

If x2y=2x-y then y2+y42+y63+……∞=

x2y=2x-y ⇒ y(1+x2)=2x ⇒ y=2x1+x2 y2+y222+y233+......∞ =-log (1-y)2 =-log 1-(2x)2(1+x2)2 =-log (1-x2)2(1+x2)2 =-2log 1-x21+x2 =2log 1+x21-x2

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If $x^{2} y = 2 x - y$ then $y^{2} + \frac{y^{4}}{2} + \frac{y^{6}}{3} + \dots \dots \infty =$

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$2 \log (\frac{1 + x}{1 - x})$

$2 \log (\frac{1 + x^{2}}{1 - x^{2}})$

$\log (\frac{1 + x}{1 - x})$

$\log (\frac{1 + x^{2}}{1 - x^{2}})$

answer is B.

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Detailed Solution

$\begin{array}{rcl} x^{2} y & = & 2 x - y \Rightarrow y (1 + x^{2}) = 2 x \end{array}$
$\begin{array}{rcl} \Rightarrow & y = \frac{2 x}{1 + x^{2}} \end{array}$
$\begin{array}{rcl} y^{2} + \frac{{(y^{2})}^{2}}{2} + \frac{{(y^{2})}^{3}}{3} + . . . . . . \infty \\ = & - \log {(1 - y)}^{2} = - \log (1 - \frac{{(2 x)}^{2}}{{(1 + x^{2})}^{2}}) \\ = & - \log (\frac{{(1 - x^{2})}^{2}}{{(1 + x^{2})}^{2}}) \\ = & - 2 \log (\frac{1 - x^{2}}{1 + x^{2}}) \\ = & 2 \log (\frac{1 + x^{2}}{1 - x^{2}}) \end{array}$