$1 + \frac{1}{4} + \frac{1 \times 3}{4 \times 8} + \frac{1 \times 3 \times 5}{4 \times 8 \times 12} + \dots = $

If $x=\frac{2}{5}+\frac{1 \cdot 3}{2 !}\left(\frac{2}{5}\right)^2+\frac{1 \cdot 3 \cdot 5}{3 !}\left(\frac{2}{5}\right)^3+\ldots$,then $x+\frac{1}{x}=$

If $x$ is a positive real number and the first negative term in the expansion of $(1+x)^{27/5}$ is $t_k$,then $k=$

If $x$ is so large that terms containing $x^{-3}, x^{-4}, x^{-5}, \ldots$ can be neglected,then the approximate value of $\left(\frac{3 x-5}{4 x^2+3}\right)^{-4 / 5}$ is

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

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