The coefficient of $x^3$ in the expansion of $(1-x)^{3/2}$,$(|x| < 1)$ is

For $x=\frac{5}{7}$,if $t_k$ is the first negative term in the expansion of $(1+x)^{7/5}$,then $t_1+t_2+\ldots+t_k=$

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

If $\alpha = \frac{5}{2! \times 3} + \frac{5 \times 7}{3! \times 3^2} + \frac{5 \times 7 \times 9}{4! \times 3^3} + \ldots$,then $\alpha^2 + 4\alpha =$

If $x = \frac{2 \cdot 5}{2! \cdot 3} + \frac{2 \cdot 5 \cdot 7}{3! \cdot 3^2} + \frac{2 \cdot 5 \cdot 7 \cdot 9}{4! \cdot 3^3} + \ldots$,then $x^2 + 8x + 8 = $

The coefficient of $x^r$ in the expansion of $\frac{1}{\sqrt[3]{(1-2 x)^2}}$ is