1+frac{2}{4}+frac{2 cdot 5}{4 cdot 8}+frac{2 | vedclass

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(B) Let $S = 1 + \frac{2}{4} + \frac{2 \cdot 5}{4 \cdot 8} + \frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12} + \ldots$ Comparing this with the binomial e

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$\sqrt[3]{16}$

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(B) Let $S = 1 + \frac{2}{4} + \frac{2 \cdot 5}{4 \cdot 8} + \frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12} + \ldots$ Comparing this with the binomial expansion $(1-x)^n = 1 + n(-x) + \frac{n(n-1)}{2!}(-x)^2 + \ldots$ Here,the general term is $\frac{2 \cdot 5 \cdot 8 \cdots (3r-1)}{4 \cdot 8 \cdot 12 \cdots (4r)}$. This is a binomial series of the form $(1-x)^n$. Comparing the terms,we have $nx = \frac{2}{4} = \frac{1}{2}$ and $\frac{n(n-1)}{2!} x^2 = \frac{2 \cdot 5}{4 \cdot 8} = \frac{10}{32} = \frac{5}{16}$. Dividing the second equation by the square of the first: $\frac{n(n-1)x^2 / 2}{n^2 x^2} = \frac{5/16}{1/4}$ $\Rightarrow \frac{n-1}{2n} = \frac{5}{4}$ $\Rightarrow 4n - 4 = 10n$ $\Rightarrow 6n = -4$ $\Rightarrow n = -2/3$. Substituting $n = -2/3$ into $nx = 1/2$,we get $(-2/3)x = 1/2 \Rightarrow x = -3/4$. Thus,$S = (1 - x)^n = (1 - (-3/4))^{-2/3} = (7/4)^{-2/3}$ is incorrect; let us re-evaluate the series form. The series is $(1-x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \ldots$. For the given series,$nx = 2/4 = 1/2$ and $\frac{n(n-1)}{2}x^2 = 10/32 = 5/16$. Using $x = 1/(2n)$,we get $\frac{n(n-1)}{2} \cdot \frac{1}{4n^2} = \frac{5}{16}$ $\Rightarrow \frac{n-1}{8n} = \frac{5}{16}$ $\Rightarrow 16n - 16 = 40n$ $\Rightarrow 24n = -16$ $\Rightarrow n = -2/3$. Then $x = 1/(2(-2/3)) = -3/4$. The sum is $(1 - (-3/4))^{-2/3} = (1 + 3/4)^{-2/3} = (7/4)^{-2/3}$? No,the series is $(1-x)^n$. Actually,the series is $(1-x)^n = 1 + n(-x) + \frac{n(n-1)}{2!}(-x)^2 + \ldots$. With $x = -3/4$ and $n = -2/3$,$S = (1 - (-3/4))^{-2/3} = (7/4)^{-2/3}$. Wait,the standard form is $(1-x)^n = 1 - nx + \frac{n(n-1)}{2}x^2 - \ldots$. Given $S = (1-x)^n = (1 - 3/4)^{-2/3} = (1/4)^{-2/3} = (4^{-1})^{-2/3} = 4^{2/3} = \sqrt[3]{16}$.

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	$(iv)$ $\frac{x}{x-1}$
	$(v)$ $\frac{x^4}{(x^2+1)^2}$
	$(vi)$ $\frac{x^4}{(x^2-1)^2}$

$1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\frac{2 \cdot 5 \cdot 8 \cdot 11}{4 \cdot 8 \cdot 12 \cdot 16}+\ldots \ldots$ is equal to :

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