1+12.35+1.32.4.925+1.3.52.4.6.27125+⋯∞=

1+12.35+1.3.92.4.25+1.3.52.4.6.27125+⋯+∞p=1,q=2(12)(35)=p1!.x2⇒310=x2⇒x=35(1−x)−pq=(1−35)−12=(25)−12=52

1+12.35+1.32.4.925+1.3.52.4.6.27125+⋯∞=

1+12.35+1.3.92.4.25+1.3.52.4.6.27125+⋯+∞p=1,q=2(12)(35)=p1!.x2⇒310=x2⇒x=35(1−x)−pq=(1−35)−12=(25)−12=52

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$1 + \frac{1}{2} . \frac{3}{5} + \frac{1.3}{2.4} . \frac{9}{25} + \frac{1.3.5}{2.4.6} . \frac{27}{125} + \dots \infty =$

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$\sqrt{\frac{5}{2}}$

$\sqrt{\frac{2}{5}}$

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

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

answer is A.

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Detailed Solution

$\begin{array}{l} 1 + \frac{1}{2} . \frac{3}{5} + \frac{1.3.9}{2.4.25} + \frac{1.3.5}{2.4.6} . \frac{27}{125} + \dots + \infty \\ p = 1, q = 2 \\ (\frac{1}{2}) (\frac{3}{5}) = \frac{p}{1!} . \frac{x}{2} \\ \Rightarrow \frac{3}{10} = \frac{x}{2} \end{array}$

$\begin{array}{l} \Rightarrow x = \frac{3}{5} \\ {(1 - x)}^{- \frac{p}{q}} = {(1 - \frac{3}{5})}^{\frac{- 1}{2}} \\ = {(\frac{2}{5})}^{- \frac{1}{2}} \\ = \sqrt{\frac{5}{2}} \end{array}$