Find a positive value of $m$ for which the coefficient of $x^{2}$ in the expansion $(1+x)^{m}$ is $6$.

If $x$ is so small that $x^2$ and higher powers of $x$ can be neglected,then the approximate value of $\left(1+\frac{3}{4} x\right)^{\frac{1}{2}}\left(1-\frac{2 x}{3}\right)^{-2}$ is

If $|x| < \frac{1}{2}$,then the coefficient of $x^r$ in the expansion of $\frac{1+2x}{(1-2x)^2}$ is

If $3x = 1 + \frac{5}{8} + \frac{5 \times 9}{8 \times 16} + \frac{5 \times 9 \times 13}{8 \times 16 \times 24} + \dots$,then $x^4 + 4x^3 + 6x^2 + 4x = $

The sum of the coefficients of $x^{-3/2}$ and $x^3$ in the expansion of $\sqrt{3+x} + \sqrt{5+x}$ when $3 < x < 5$ is

$1+\frac{1}{3}+\frac{1 \times 3}{3 \times 6}+\frac{1 \times 3 \times 5}{3 \times 6 \times 9}+\ldots \text{ to } \infty =$