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

The sum of the series $\frac{3}{4 \cdot 8}-\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}-\ldots$

$1 - \frac{3}{16} + \frac{1 \cdot 4}{1 \cdot 2} \left(\frac{3}{16}\right)^2 - \frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3} \left(\frac{3}{16}\right)^3 + \ldots$

If $|x| < 1$,then the number of terms in the expansion of $[\frac{1}{2}(1 \cdot 2 + 2 \cdot 3 x + 3 \cdot 4 x^2 + . . . . . . \infty)]^{-25}$ is

The ${(r + 1)^{th}}$ term in the expansion of ${(1 - x)^{-4}}$ is:

If $y = \frac{3}{4} + \frac{3 \times 5}{4 \times 8} + \frac{3 \times 5 \times 7}{4 \times 8 \times 12} + \ldots$ to $\infty$,then