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

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(B) Given the series $y = \frac{3}{4} + \frac{3 	imes 5}{4 	imes 8} + \frac{3 	imes 5 	imes 7}{4 	imes 8 	imes 12} + \ldots \infty$. Adding $1$

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

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(B) Given the series $y = \frac{3}{4} + \frac{3 	imes 5}{4 	imes 8} + \frac{3 	imes 5 	imes 7}{4 	imes 8 	imes 12} + \ldots \infty$. Adding $1$ to both sides,we get $y + 1 = 1 + \frac{3}{4} + \frac{3 	imes 5}{4 	imes 8} + \frac{3 	imes 5 	imes 7}{4 	imes 8 	imes 12} + \ldots \infty$. This is of the form $(1 - x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \ldots$. Here,$nx = \frac{3}{4}$ and $\frac{n(n+1)}{2!}x^2 = \frac{15}{32}$. Solving for $n$ and $x$,we find $n = \frac{3}{2}$ and $x = \frac{1}{2}$. Thus,$y + 1 = (1 - \frac{1}{2})^{-\frac{3}{2}} = (\frac{1}{2})^{-\frac{3}{2}} = 2^{\frac{3}{2}}$. Squaring both sides,$(y + 1)^2 = 2^3 = 8$. $y^2 + 2y + 1 = 8 \Rightarrow y^2 + 2y - 7 = 0$.

If $y = \frac{3}{4} + \frac{3 \times 5}{4 \times 8} + \frac{3 \times 5 \times 7}{4 \times 8 \times 12} + \ldots$ to $\infty$,then

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