frac{1}{8} - frac{7}{8 times 12} + frac{7 tim | vedclass

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Question

(B) Let $S = \frac{1}{8} - \frac{7}{8 	imes 12} + \frac{7 	imes 10}{8 	imes 12 	imes 16} - \ldots$ Using the binomial expansion $(1+x)^n = 1 + nx

Vedclass · Accepted Answer

$\sqrt[3]{\frac{4}{7}} - \frac{3}{4}$

Vedclass · Answer

(B) Let $S = \frac{1}{8} - \frac{7}{8 	imes 12} + \frac{7 	imes 10}{8 	imes 12 	imes 16} - \ldots$ Using the binomial expansion $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots$ Consider $(1 + \frac{3}{4})^{-1/3} = 1 + (-\frac{1}{3})(\frac{3}{4}) + \frac{(-\frac{1}{3})(-\frac{4}{3})}{2!}(\frac{3}{4})^2 + \frac{(-\frac{1}{3})(-\frac{4}{3})(-\frac{7}{3})}{3!}(\frac{3}{4})^3 + \ldots$ $= 1 - \frac{1}{4} + \frac{4}{18} 	imes \frac{9}{16} - \frac{28}{162} 	imes \frac{27}{64} + \ldots$ $= 1 - \frac{1}{4} + \frac{1}{8} - \frac{7}{8 	imes 12} + \ldots$ $= \frac{3}{4} + S$ Since $(1 + \frac{3}{4})^{-1/3} = (\frac{7}{4})^{-1/3} = \sqrt[3]{\frac{4}{7}}$,we have $\sqrt[3]{\frac{4}{7}} = \frac{3}{4} + S$ Therefore,$S = \sqrt[3]{\frac{4}{7}} - \frac{3}{4}$

$\frac{1}{8} - \frac{7}{8 \times 12} + \frac{7 \times 10}{8 \times 12 \times 16} - \ldots =$

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