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

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Question

(B) Let the 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} - \ldots

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the series be $S = \frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - \ldots$ We know the binomial expansion $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots$ For $n = -1/2$,$(1+x)^{-1/2} = 1 - \frac{1}{2}x + \frac{(-1/2)(-3/2)}{2}x^2 - \frac{(-1/2)(-3/2)(-5/2)}{6}x^3 + \ldots = 1 - \frac{1}{2}x + \frac{1 \cdot 3}{2 \cdot 4}x^2 - \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}x^3 + \ldots$ Multiply the given series by $2$ and adjust terms to match the form: $S = \frac{1}{2} \left[ \frac{3 \cdot 2}{4 \cdot 8} - \frac{3 \cdot 5 \cdot 2}{4 \cdot 8 \cdot 12} + \ldots \right] = \frac{1}{2} \left[ \frac{1 \cdot 3}{4 \cdot 4} - \frac{1 \cdot 3 \cdot 5}{4 \cdot 4 \cdot 6} + \ldots \right]$ Using the expansion with $x = 1/2$,we find the sum is $\sqrt{\frac{2}{3}} - \frac{3}{4}$.

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

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