To expand $(1 + 2x)^{-1/2}$ as an infinite series,the range of $x$ should be

If $x=1+\frac{3}{1!} \times \frac{1}{6}+\frac{3 \times 7}{2!}\left(\frac{1}{6}\right)^2+\frac{3 \times 7 \times 11}{3!}\left(\frac{1}{6}\right)^3+\ldots$,then $x^4$ equals

The expansion of $(1+x+x^2)^{-3/2}$ in powers of $x$ is valid if

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

If $(2-5x)^{-1/5} = a_0 + a_1x + a_2x^2 + \ldots$,then $\frac{a_1}{a_2} = $

If $y = \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \dots \infty$,then