The coefficient of $x^3$ in the expansion of $\frac{(1 + 3x)^2}{1 - 2x}$ is:

The sum of the series $\frac{3}{4 \cdot 8}-\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}-\ldots$

For $|x| < \frac{4}{3}$, the approximate value of $\frac{1}{(4-3 x)^{\frac{1}{2}}}$ is

If $x = \frac{3}{4 \cdot 8} + \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} + \ldots$,then $2x^2 + 5x =$

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

If $x$ is small,so that $x^2$ and higher powers can be neglected,then the approximate value for $\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}}$ is