The sum of the series $1 + \frac{1 \times 3}{6} + \frac{1 \times 3 \times 5}{6 \times 8} + \dots \infty$ is

The cube root of $217$ is:

If $(-c, c)$ is the set of all values of $x$ for which the expansion of $(7-5x)^{-2/3}$ is valid,then $5c + 7 =$

If the value of $x$ is so small that $x^2$ and higher powers can be neglected,then $\frac{\sqrt{1 + x} + \sqrt[3]{(1 - x)^2}}{1 + x + \sqrt{1 + x}}$ is equal to

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

The correct matching of List-$I$ from List-$II$ is:

List-$I$	List-$II$
$(A)$ $(1-x)^{-n}$	$(i)$ $\frac{x}{x+1}$
$(B)$ $(1+x)^{-n}$	$(ii)$ $1-nx+\frac{n(n+1)}{2!}x^2-\dots$ if $|x| < 1$
$(C)$ If $x>1$,then $1+\frac{1}{x}+\frac{1}{x^2}+\dots$ is	$(iii)$ $1+nx+\frac{n(n+1)}{2!}x^2+\dots$ if $|x| < 1$
$(D)$ If $|x|>1$,then $1-\frac{2}{x^2}+\frac{3}{x^4}-\frac{4}{x^6}+\dots$ is	$(iv)$ $\frac{x}{x-1}$
	$(v)$ $\frac{x^4}{(x^2+1)^2}$
	$(vi)$ $\frac{x^4}{(x^2-1)^2}$