The sum of the series 1+frac{2}{3}left(frac{1 | vedclass

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(A) The given series is of the form $1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \dots = (1-x)^{-n}$.Comparing the terms,we have $nx =

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$\frac{4}{\sqrt[3]{49}}$

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(A) The given series is of the form $1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \dots = (1-x)^{-n}$.Comparing the terms,we have $nx = \frac{2}{3} 	imes \frac{1}{8} = \frac{1}{12}$ and $x = \frac{1}{8}$.Thus,$n 	imes \frac{1}{8} = \frac{1}{12} \implies n = \frac{8}{12} = \frac{2}{3}$.The series is $(1 - \frac{1}{8})^{-\frac{2}{3}}$.$= (\frac{7}{8})^{-\frac{2}{3}} = (\frac{8}{7})^{\frac{2}{3}} = \frac{8^{\frac{2}{3}}}{7^{\frac{2}{3}}} = \frac{(2^3)^{\frac{2}{3}}}{7^{\frac{2}{3}}} = \frac{2^2}{\sqrt[3]{49}} = \frac{4}{\sqrt[3]{49}}$.

The sum of the series $1+\frac{2}{3}\left(\frac{1}{8}\right)+\frac{2 \times 5}{3 \times 6}\left(\frac{1}{8}\right)^2+\frac{2 \times 5 \times 8}{3 \times 6 \times 9}\left(\frac{1}{8}\right)^3+\ldots$ is

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