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

If $x = \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12} + \ldots$ to infinite terms,then $9x^2 + 24x = $

If $|x| > 1$,then $(1 + x)^{-2} = $

If $n$ is a positive integer and the coefficient of $x^{10}$ in the expansion of $(1+x)^{15}$ is equal to the coefficient of $x^5$ in the expansion of $(1-x)^{-n}$,then $n=$

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

Assuming $x$ to be so small that $x^2$ and higher powers of $x$ can be neglected,the coefficient of $x$ in $\frac{(1-x)^{1/3}+(1-5x)^2}{(16-x)^{1/4}}$ is equal to