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

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(B) The given series is $y = \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \dots \infty$. Adding $1$ to b

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$y^2 + 2y - 7 = 0$

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(B) The given series is $y = \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \dots \infty$. Adding $1$ to both sides,we get $1 + y = 1 + \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \dots \infty$. This is of the form $(1 - x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \dots$. Comparing terms,we have $nx = \frac{3}{4}$ and $\frac{n(n+1)}{2}x^2 = \frac{3 \cdot 5}{4 \cdot 8} = \frac{15}{32}$. From $nx = \frac{3}{4}$,we get $x = \frac{3}{4n}$. Substituting into the second term: $\frac{n(n+1)}{2} \cdot \frac{9}{16n^2} = \frac{15}{32} \implies \frac{n+1}{n} \cdot \frac{9}{32} = \frac{15}{32} \implies \frac{n+1}{n} = \frac{15}{9} = \frac{5}{3}$. Solving for $n$: $3n + 3 = 5n \implies 2n = 3 \implies n = \frac{3}{2}$. Then $x = \frac{3}{4 \cdot (3/2)} = \frac{3}{6} = \frac{1}{2}$. Thus,$1 + y = (1 - 1/2)^{-3/2} = (1/2)^{-3/2} = 2^{3/2} = 2\sqrt{2}$. Squaring both sides: $(1 + y)^2 = (2\sqrt{2})^2 = 8$. $1 + 2y + y^2 = 8 \implies y^2 + 2y - 7 = 0$.

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 $y = \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \dots \infty$,then

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