If x = frac{3}{4 cdot 8} + frac{3 cdot 5}{4 c | vedclass

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(A) The given series is $x = \sum_{n=2}^{\infty} \frac{3 \cdot 5 \cdot \ldots \cdot (2n+1)}{4 \cdot 8 \cdot \ldots \cdot 4n}$.This can be rewritten as

Vedclass · Accepted Answer

$\frac{7}{8}$

Vedclass · Answer

(A) The given series is $x = \sum_{n=2}^{\infty} \frac{3 \cdot 5 \cdot \ldots \cdot (2n+1)}{4 \cdot 8 \cdot \ldots \cdot 4n}$.This can be rewritten as $x = \sum_{n=2}^{\infty} \frac{\frac{1}{2} \cdot \frac{3}{2} \cdot \ldots \cdot \frac{2n+1}{2}}{n!} \left(\frac{1}{2}\right)^n$.Using the binomial expansion $(1-y)^{-n} = 1 + ny + \frac{n(n+1)}{2!}y^2 + \ldots$,we identify the series as part of $(1-y)^{-1/2}$.Specifically,$x = (1 - 1/2)^{-1/2} - 1 - (-1/2)(1/2) = \sqrt{2} - 1 - 1/4 = \sqrt{2} - 5/4$.Now,calculate $2x^2 + 5x$:$2x^2 = 2(\sqrt{2} - 5/4)^2 = 2(2 + 25/16 - 5\sqrt{2}/2) = 4 + 25/8 - 5\sqrt{2}$.$5x = 5(\sqrt{2} - 5/4) = 5\sqrt{2} - 25/4$.Adding these: $2x^2 + 5x = 4 + 25/8 - 5\sqrt{2} + 5\sqrt{2} - 50/8 = 4 - 25/8 = 7/8$.

List-$I$	List-$II$
$(A)$ $(1-x)^{-n}$	$(i)$ $\frac{x}{x+1}$
$(B)$ $(1+x)^{-n}$	$(ii)$ $1-nx+\frac{n(n+1)}{2!}x^2-\dots$ if $\|x\| < 1$
$(C)$ If $x>1$,then $1+\frac{1}{x}+\frac{1}{x^2}+\dots$ is	$(iii)$ $1+nx+\frac{n(n+1)}{2!}x^2+\dots$ if $\|x\| < 1$
$(D)$ If $\|x\|>1$,then $1-\frac{2}{x^2}+\frac{3}{x^4}-\frac{4}{x^6}+\dots$ is	$(iv)$ $\frac{x}{x-1}$
	$(v)$ $\frac{x^4}{(x^2+1)^2}$
	$(vi)$ $\frac{x^4}{(x^2-1)^2}$

If $x = \frac{3}{4 \cdot 8} + \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} + \ldots$,then $2x^2 + 5x =$

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