The fourth term in the expansion of $(1 - 2x)^{3/2}$ is:

The coefficient of $x^3$ in the expansion of $(1-\frac{3}{4} x)^{\frac{1}{2}}$ is

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

If $x > \sqrt{3}$ and $\frac{x^2+1}{(x^2+2)(x^2+3)}$ is expanded in terms of powers of $x^{-2}$,then the coefficient of $x^{-8}$ is

If $|x| > 1$,then $(1 + x)^{-2} = $

Assertion $(A) : 1+\frac{2}{3} \cdot \frac{1}{2}+\frac{2 \cdot 5}{3 \cdot 6} \cdot \frac{1}{4}+\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9} \cdot \frac{1}{8}+\ldots \infty = \sqrt[3]{4}$
Reason $(R) : |x| < 1, (1-x)^{-n} = 1+nx+\frac{n(n+1)}{1 \cdot 2} x^2+\frac{n(n+1)(n+2)}{1 \cdot 2 \cdot 3} x^3+\ldots$ The correct answer is