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

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

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

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(B) Given the series: $y = \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \ldots \infty$. Adding $1$ to both sides: $1 + y = 1 + \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \ldots \infty$. Using the binomial expansion $(1 - x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \ldots$,where $n = \frac{3}{2}$ and $x = \frac{1}{2}$,we have: $1 + y = (1 - \frac{1}{2})^{-\frac{3}{2}} = (\frac{1}{2})^{-\frac{3}{2}} = 2^{\frac{3}{2}} = \sqrt{8} = 2\sqrt{2}$. Squaring both sides: $(1 + y)^2 = (2\sqrt{2})^2$. $1 + y^2 + 2y = 8$. $y^2 + 2y - 7 = 0$.

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

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