If x = frac{1 cdot 3}{3 cdot 6} + frac{1 cdot | vedclass

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(B) Given that,$x = \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cd

Vedclass · Accepted Answer

$11$

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(B) Given that,$x = \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12} + \ldots \infty 	ext{ terms}$.We can rewrite the terms as:$x = \frac{1 \cdot 3}{3^2 \cdot 2!} + \frac{1 \cdot 3 \cdot 5}{3^3 \cdot 3!} + \frac{1 \cdot 3 \cdot 5 \cdot 7}{3^4 \cdot 4!} + \ldots$Using the binomial expansion $(1-z)^{-n} = 1 + nz + \frac{n(n+1)}{2!}z^2 + \ldots$,we identify $n = 1/2$ and $z = 2/3$.$x = \left[ 1 + \frac{1}{2}(\frac{2}{3}) + \frac{\frac{1}{2}(\frac{1}{2}+1)}{2!}(\frac{2}{3})^2 + \ldots ight] - (1 + \frac{1}{3})$$x = (1 - \frac{2}{3})^{-1/2} - \frac{4}{3}$$x = (\frac{1}{3})^{-1/2} - \frac{4}{3} = \sqrt{3} - \frac{4}{3}$$3x + 4 = 3\sqrt{3}$Squaring both sides:$(3x + 4)^2 = (3\sqrt{3})^2$$9x^2 + 24x + 16 = 27$$9x^2 + 24x = 11$.

If $x = \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12} + \ldots$ to infinite terms,then $9x^2 + 24x = $

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