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

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

When $|x| < \frac{1}{2}$,the coefficient of $x^6$ in the expansion of $\left(\frac{2-x}{1+2x}\right)^2$ is

$1 + \frac{1}{3}x + \frac{1 \cdot 4}{3 \cdot 6}x^2 + \frac{1 \cdot 4 \cdot 7}{3 \cdot 6 \cdot 9}x^3 + \dots$ is equal to

If $|x| < 1$,then in the expansion of $(1 + 2x + 3x^2 + 4x^3 + ....)^{1/2}$,the coefficient of $x^n$ is

If $T_4$ represents the $4^{th}$ term in the expansion of $\left(5x + \frac{7}{x}\right)^{-3/2}$ and $x \notin \left[-\sqrt{\frac{7}{5}}, \sqrt{\frac{7}{5}}\right]$,then $\left(x^7 \sqrt{5x}\right) T_4 =$