The coefficient of $x^n$, where $n$ is any positive integer, in the expansion of $(1+2x+3x^2+\ldots)^{-1/2}$ is

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

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

If $\frac{(1 - 3x)^{1/2} + (1 - x)^{5/3}}{\sqrt{4 - x}}$ is approximately equal to $a + bx$ for small values of $x$,then $(a,b) = $

The sum of the series $\frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - \dots$ is:

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