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

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 $x+\frac{1}{x}=$

Assuming $|x|$ to be so small that $x^2$ and higher powers of $x$ can be neglected,then $\frac{\sqrt{1+x}+(1-x)^{3/2}}{(1+x)+\sqrt{1+x}} = $

The expression $\frac{1}{\sqrt{5 + 4x}}$ can be expanded by the binomial theorem,if

If $(2-5x)^{-1/5} = a_0 + a_1x + a_2x^2 + \ldots$,then $\frac{a_1}{a_2} = $

When $|x|>3$,the coefficient of $\frac{1}{x^n}$ in the expansion of $x^{3/2}(3+x)^{1/2}$ is