If x=1+frac{3}{1!} times frac{1}{6}+frac{3 ti | vedclass

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(C) The given series is of the form $(1-\alpha)^{-p/q} = 1 + \frac{p}{1!}(\frac{\alpha}{q}) + \frac{p(p+q)}{2!}(\frac{\alpha}{q})^2 + \frac{p(p+q)(p+2

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$27$

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(C) The given series is of the form $(1-\alpha)^{-p/q} = 1 + \frac{p}{1!}(\frac{\alpha}{q}) + \frac{p(p+q)}{2!}(\frac{\alpha}{q})^2 + \frac{p(p+q)(p+2q)}{3!}(\frac{\alpha}{q})^3 + \ldots$ Comparing the given series with this expansion,we have $p=3$,$p+q=7$,and $p+2q=11$. From $p=3$ and $p+q=7$,we get $q=4$. Also,$\frac{\alpha}{q} = \frac{1}{6}$,so $\alpha = \frac{q}{6} = \frac{4}{6} = \frac{2}{3}$. Thus,$x = (1-\alpha)^{-p/q} = (1-\frac{2}{3})^{-3/4} = (\frac{1}{3})^{-3/4} = (3)^{3/4}$. Therefore,$x^4 = (3^{3/4})^4 = 3^3 = 27$.

If $x=1+\frac{3}{1!} \times \frac{1}{6}+\frac{3 \times 7}{2!}\left(\frac{1}{6}\right)^2+\frac{3 \times 7 \times 11}{3!}\left(\frac{1}{6}\right)^3+\ldots$,then $x^4$ equals

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