1−15+1.45.10−1.4.75.10.15+...∞=

1−15+1.45.10−1.4.75.10.15+...∞1−p1!(xq)+p(p+q)2!(xq)2−...p=1 , q=3 ,xq=15⇒x=35⇒(1+35)−13=(85)−13=12(513)=1253

1−15+1.45.10−1.4.75.10.15+...∞=

1−15+1.45.10−1.4.75.10.15+...∞1−p1!(xq)+p(p+q)2!(xq)2−...p=1 , q=3 ,xq=15⇒x=35⇒(1+35)−13=(85)−13=12(513)=1253

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$1 - \frac{1}{5} + \frac{1.4}{5.10} - \frac{1.4.7}{5.10.15} + ... \infty =$

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$\frac{1}{3} \sqrt[3]{5}$

$\frac{1}{2} \sqrt[3]{5}$

$\frac{1}{2} \sqrt[3]{4}$

$\sqrt[3]{5}$

answer is B.

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$1 - \frac{1}{5} + \frac{1.4}{5.10} - \frac{1.4.7}{5.10.15} + ... \infty$

$1 - \frac{p}{1!} (\frac{x}{q}) + \frac{p (p + q)}{2!} {(\frac{x}{q})}^{2} - ...$

$p = 1, q = 3, \frac{x}{q} = \frac{1}{5} \Rightarrow x = \frac{3}{5}$

$\Rightarrow {(1 + \frac{3}{5})}^{- \frac{1}{3}} = {(\frac{8}{5})}^{- \frac{1}{3}} = \frac{1}{2} (5^{\frac{1}{3}})$

$= \frac{1}{2} \sqrt[3]{5}$

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1−15+1.45.10−1.4.75.10.15+...∞=

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$1 - \frac{1}{5} + \frac{1.4}{5.10} - \frac{1.4.7}{5.10.15} + ... \infty =$