If x=frac{2}{5}+frac{1 cdot 3}{2 !}left(frac{ | vedclass

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(D) The given series is of the form $(1-y)^{-n} - 1 = ny + \frac{n(n+1)}{2!}y^2 + \frac{n(n+1)(n+2)}{3!}y^3 + \ldots$ Comparing the given series $x =

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$\frac{5 \sqrt{5}-3}{4}$

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(D) The given series is of the form $(1-y)^{-n} - 1 = ny + \frac{n(n+1)}{2!}y^2 + \frac{n(n+1)(n+2)}{3!}y^3 + \ldots$ Comparing the given series $x = \frac{2}{5} + \frac{1 \cdot 3}{2!}(\frac{2}{5})^2 + \frac{1 \cdot 3 \cdot 5}{3!}(\frac{2}{5})^3 + \ldots$ with the binomial expansion,we identify $ny = \frac{2}{5}$ and $y = -\frac{2}{5}$ (since the terms are positive). For the series $\frac{1 \cdot 3 \cdot 5 \ldots (2n-1)}{n!} y^n$,we have $n = -1/2$ and $y = -2/5$. Thus,$x = (1 - (-2/5))^{-1/2} - 1 = (7/5)^{-1/2} - 1$. Wait,let us re-evaluate: The series is $(1-y)^{-n} - 1$. For $(1-y)^{-1/2} = 1 + \frac{1}{2}y + \frac{1 \cdot 3}{2 \cdot 4}y^2 + \ldots$. Using the expansion $(1-y)^{-1/2} - 1 = \frac{1}{2}y + \frac{1 \cdot 3}{2 \cdot 4}y^2 + \ldots$. Comparing $x = \frac{2}{5} + \frac{1 \cdot 3}{2!}(\frac{2}{5})^2 + \ldots$,we find $y = 4/5$ and $n = 1/2$. $x = (1 - 4/5)^{-1/2} - 1 = (1/5)^{-1/2} - 1 = \sqrt{5} - 1$. Then $\frac{1}{x} = \frac{1}{\sqrt{5}-1} = \frac{\sqrt{5}+1}{4}$. $x + \frac{1}{x} = \sqrt{5} - 1 + \frac{\sqrt{5}+1}{4} = \frac{4\sqrt{5} - 4 + \sqrt{5} + 1}{4} = \frac{5\sqrt{5} - 3}{4}$. Therefore,the correct option is $D$.

If $x=\frac{2}{5}+\frac{1 \cdot 3}{2 !}\left(\frac{2}{5}\right)^2+\frac{1 \cdot 3 \cdot 5}{3 !}\left(\frac{2}{5}\right)^3+\ldots$,then $x+\frac{1}{x}=$

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