If $x = \frac{1}{5} + \frac{1 \times 3}{5 \times 10} + \frac{1 \times 3 \times 5}{5 \times 10 \times 15} + \ldots$,then $3x^2 + 6x =$

$\frac{1}{8} - \frac{7}{8 \times 12} + \frac{7 \times 10}{8 \times 12 \times 16} - \ldots =$

The sum of the infinite series $1+\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots$ is equal to

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

If $x$ is numerically so small that $x^2$ and higher powers of $x$ can be neglected, then $\left(1+\frac{2x}{3}\right)^{3/2} \cdot (32+5x)^{-1/5}$ is approximately equal to

If $|x| < 1$,then the number of terms in the expansion of $[\frac{1}{2}(1 \cdot 2 + 2 \cdot 3 x + 3 \cdot 4 x^2 + . . . . . . \infty)]^{-25}$ is