The $13^{\text{th}}$ term in the expansion of $(1-4x)^{-4}$ is

$1 - \frac{1}{8} + \frac{1}{8} \cdot \frac{3}{16} - \frac{1 \cdot 3 \cdot 5}{8 \cdot 16 \cdot 24} + \dots =$

For $0 < x < 1$,the expansion of $\left(1+\frac{1}{x}\right)^{\frac{1}{2}}$ is

The first negative coefficient in the terms occurring in the expansion of $(1+x)^{\frac{21}{5}}$ is

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 $7^2(12 x+55)^3=$

If $\alpha = \frac{5}{2! \times 3} + \frac{5 \times 7}{3! \times 3^2} + \frac{5 \times 7 \times 9}{4! \times 3^3} + \ldots$,then $\alpha^2 + 4\alpha =$