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

The sum of the series $1+\frac{2}{3}\left(\frac{1}{8}\right)+\frac{2 \times 5}{3 \times 6}\left(\frac{1}{8}\right)^2+\frac{2 \times 5 \times 8}{3 \times 6 \times 9}\left(\frac{1}{8}\right)^3+\ldots$ 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

Find the binomial expansion of $\frac{1}{\sqrt[3]{6 - 3x}}$.

The set of all real values of $x$ for which the expansion of $\left(125 x^2-\frac{27}{x}\right)^{-2/3}$ is valid,is