frac{1}{4}-frac{5}{4 cdot 8}+frac{5 cdot 9}{4 | vedclass

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(C) Let the given series be $S = \frac{1}{4} - \frac{5}{4 \cdot 8} + \frac{5 \cdot 9}{4 \cdot 8 \cdot 12} - \ldots$This is a binomial series of the fo

Vedclass · Accepted Answer

$\frac{2^{\frac{1}{4}}-1}{2^{\frac{1}{4}}}$

Vedclass · Answer

(C) Let the given series be $S = \frac{1}{4} - \frac{5}{4 \cdot 8} + \frac{5 \cdot 9}{4 \cdot 8 \cdot 12} - \ldots$This is a binomial series of the form $(1+x)^{-n} = 1 - nx + \frac{n(n+1)}{2!}x^2 - \frac{n(n+1)(n+2)}{3!}x^3 + \ldots$We can rewrite the series as $S = 1 - [1 - \frac{1}{4} + \frac{1 \cdot 5}{2! \cdot 4^2} - \frac{1 \cdot 5 \cdot 9}{3! \cdot 4^3} + \ldots]$Comparing the terms inside the bracket with the binomial expansion $(1+x)^{-n}$,we have $nx = \frac{1}{4}$ and $n = \frac{1}{4}$,which gives $x = 1$.Thus,the series inside the bracket is $(1+1)^{-\frac{1}{4}} = 2^{-\frac{1}{4}}$.Therefore,$S = 1 - 2^{-\frac{1}{4}} = 1 - \frac{1}{2^{\frac{1}{4}}} = \frac{2^{\frac{1}{4}}-1}{2^{\frac{1}{4}}}$.

$\frac{1}{4}-\frac{5}{4 \cdot 8}+\frac{5 \cdot 9}{4 \cdot 8 \cdot 12}-\ldots=$

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