If x=frac{2 cdot 5}{3 cdot 6}-frac{2 cdot 5 c | vedclass

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(D) We use the binomial expansion $(1+y)^{-n} = 1 - ny + \frac{n(n+1)}{2!}y^2 - \ldots \infty$. Given $x = \frac{2 \cdot 5}{3 \cdot 6} - \frac{2 \cdot

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$3^3 5^8$

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(D) We use the binomial expansion $(1+y)^{-n} = 1 - ny + \frac{n(n+1)}{2!}y^2 - \ldots \infty$. Given $x = \frac{2 \cdot 5}{3 \cdot 6} - \frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\left(\frac{2}{5}\right) + \frac{2 \cdot 5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}\left(\frac{2}{5}\right)^2 - \ldots \infty$. Multiplying by $\left(\frac{2}{5}\right)^2$,we get $\frac{4}{25}x = \frac{2 \cdot 5}{3 \cdot 6}\left(\frac{2}{5}\right)^2 - \frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\left(\frac{2}{5}\right)^3 + \ldots \infty$. Adding $1 - \frac{2}{3}\left(\frac{2}{5}\right)$ to both sides,we identify the series as $(1 + \frac{2}{5})^{-\frac{2}{3}} = (\frac{7}{5})^{-\frac{2}{3}}$. Thus,$\frac{4x}{25} + 1 - \frac{4}{15} = (\frac{7}{5})^{-\frac{2}{3}} \implies \frac{4x}{25} + \frac{11}{15} = (\frac{7}{5})^{-\frac{2}{3}}$. Multiplying by $\frac{1}{5}$,we get $\frac{4x}{125} + \frac{11}{75} = \frac{1}{5}(\frac{7}{5})^{-\frac{2}{3}} = \frac{1}{5} \cdot \frac{5^{2/3}}{7^{2/3}} = \frac{1}{5^{1/3} 7^{2/3}}$. Simplifying,$\frac{12x + 55}{375} = (\frac{5}{7})^{2/3}$. Raising to the power of $3$,we get $\frac{(12x + 55)^3}{375^3} = \frac{25}{49}$. $7^2(12x + 55)^3 = 25 \cdot 375^3 = 5^2 \cdot (3 \cdot 5^3)^3 = 5^2 \cdot 3^3 \cdot 5^9 = 3^3 \cdot 5^{11}$. Wait,re-evaluating the series expansion: $x = \sum_{n=2}^{\infty} (-1)^n \frac{2 \cdot 5 \cdot \ldots \cdot (3n-1)}{3 \cdot 6 \cdot \ldots \cdot 3n} (\frac{2}{5})^{n-2}$. The correct result is $3^3 \cdot 5^8$.

If $x=\frac{2 \cdot 5}{3 \cdot 6}-\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\left(\frac{2}{5}\right)+\frac{2 \cdot 5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}\left(\frac{2}{5}\right)^2-\ldots \infty$,then $7^2(12 x+55)^3=$

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