1 - frac{1}{8} + frac{1}{8} cdot frac{3}{16} | vedclass

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(C) The given series is $1 - \frac{1}{8} + \frac{1 \cdot 3}{8 \cdot 16} - \frac{1 \cdot 3 \cdot 5}{8 \cdot 16 \cdot 24} + \dots$Comparing this with th

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

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(C) The given series is $1 - \frac{1}{8} + \frac{1 \cdot 3}{8 \cdot 16} - \frac{1 \cdot 3 \cdot 5}{8 \cdot 16 \cdot 24} + \dots$Comparing this with the binomial expansion $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$,we have:$nx = -\frac{1}{8}$  $(1)$$\frac{n(n-1)}{2}x^2 = \frac{3}{128}$  $(2)$From $(1)$,$x = -\frac{1}{8n}$. Substituting this into $(2)$:$\frac{n(n-1)}{2} \cdot \frac{1}{64n^2} = \frac{3}{128}$$\frac{n-1}{128n} = \frac{3}{128}$ $\Rightarrow n-1 = 3n$ $\Rightarrow 2n = -1$ $\Rightarrow n = -\frac{1}{2}$Substituting $n = -\frac{1}{2}$ into $(1)$:$-\frac{1}{2}x = -\frac{1}{8} \Rightarrow x = \frac{1}{4}$Thus,the sum is $(1 + x)^n = (1 + \frac{1}{4})^{-1/2} = (\frac{5}{4})^{-1/2} = (\frac{4}{5})^{1/2} = \frac{2}{\sqrt{5}}$.

$1 - \frac{1}{8} + \frac{1}{8} \cdot \frac{3}{16} - \frac{1 \cdot 3 \cdot 5}{8 \cdot 16 \cdot 24} + \dots =$

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