Explanation

(b)
$Let\ S = 1 + \frac{2}{4} + \frac{2 \bullet 5}{4 \bullet 8} + \frac{2 \bullet 5 \bullet 8}{4 \bullet 8 \bullet 12} + \ldots$
On comparing with
$(1 + x)^{n} = 1 + nx + \frac{n(n - 1)}{2!}x^{2} + \ldots$
$we\ get\ \ \ \ \ \ \ \ \ \ \ \ \ \ nx = \frac{2}{4}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
$and\ \ \ \ \frac{n(n - 1)}{2!}x^{2} = \frac{2 \bullet 5}{4 \bullet 8}\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eqs, (i) and (ii)
$\ \ \ \ \ \ \ \ \ \ \ \frac{\frac{n(n - 1)}{2!}x^{2}}{x^{2}x^{2}} = \frac{\frac{2 \bullet 5}{4 \bullet 8}}{\frac{2 \bullet 2}{4 \bullet 4}}$
$\Longrightarrow \frac{n - 1}{n} = \frac{5}{2} \Longrightarrow n = - \frac{2}{3}$
On putting the value of $n$ in Eq. (i) we get
$\ \ \ \ \ \ \ \ \ \ \ \ - \frac{2}{3}x = \frac{2}{4} \Longrightarrow x = - \frac{3}{4}$
$\therefore S = {(1 + x)}^{n} = \left( 1 - \frac{3}{4} \right)^{- 2/3} = \left( \frac{1}{4} \right)^{- 2/3\ } = \sqrt[3]{16}$