The sum of the series frac{3}{4 cdot 8} - fra | vedclass

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Question

(B) Let the given series be $S = \frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} -

Vedclass · Accepted Answer

$\sqrt{\frac{2}{3}} - \frac{3}{4}$

Vedclass · Answer

(B) Let the given series be $S = \frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - \dots$ We know the binomial expansion $(1+x)^{-n} = 1 - nx + \frac{n(n+1)}{2!}x^2 - \dots$ Adding and subtracting $\frac{3}{4}$ to the series,we get: $S = \frac{3}{4} + \frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \dots - \frac{3}{4}$ $S = 1 - \frac{1}{4} + \frac{1 \cdot 3}{2! \cdot 4^2} - \frac{1 \cdot 3 \cdot 5}{3! \cdot 4^3} + \dots - \frac{3}{4}$ This is the expansion of $(1 + \frac{1}{4})^{-1/2} - \frac{3}{4}$ $S = (\frac{5}{4})^{-1/2} - \frac{3}{4} = \sqrt{\frac{4}{5}} - \frac{3}{4}$ Wait,re-evaluating the series expansion: $S = \frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \dots$ Using the form $(1+x)^{-1/2} = 1 - \frac{1}{2}x + \frac{1 \cdot 3}{2 \cdot 4}x^2 - \dots$ Setting $x = \frac{1}{2}$,we get the sum as $\sqrt{\frac{2}{3}} - \frac{3}{4}$.

The sum of the series $\frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - \dots$ is:

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