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

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}{4} + \frac{1 \times 3}{4 \times 8} + \frac{1 \times 3 \times 5}{4 \times 8 \times 12} + \dots = $

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

If $x = \frac{1}{5} + \frac{1 \times 3}{5 \times 10} + \frac{1 \times 3 \times 5}{5 \times 10 \times 15} + \ldots$,then $3x^2 + 6x =$

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 = $