Explanation

(b)
Let, $S = 1 + \frac{1}{3} \bullet \frac{1}{2} + \frac{2}{3} \bullet \frac{5}{6} \bullet \frac{1}{2^{2}} + \frac{2}{3} \bullet \frac{5}{6} \bullet \frac{8}{9} \bullet \frac{1}{2^{3}} + ..\ldots\infty\$
and we know that
$(1 + x)^{n} = 1 + nx + \frac{n(n - 1)}{2!}x^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \frac{(n(n - 1)(n - 2)}{3!}x^{2} + \ldots\infty$
On comparing these two, we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ nx = \frac{2}{3} \bullet \frac{1}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
$and\ \frac{n(n - 1)}{2 \bullet 1}x^{2} = \frac{2}{3} \bullet \frac{5}{6} \bullet \frac{1}{2^{2}}\ \ \ \ \ \ \ \ \ldots(ii)$
from Eqs. (i) and (ii),
$\Longrightarrow \ \ \ \ \ \ \frac{\frac{n(n - 1)}{2 \bullet 1}}{n^{2}} = \frac{\frac{2}{3} \times \frac{5}{6} \times \frac{1}{4}}{\frac{2}{3} \times \frac{1}{2} \times \frac{2}{3} \times \frac{1}{2}}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n - 1}{2n} = \frac{5}{4}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5n = 2n - 2$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = - \frac{2}{3}$
On putting value of $n$ in Eq. (i), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = - \frac{1}{2}$
$\therefore Sum\ of\ series = \left( 1 - \frac{1}{2} \right)^{- \frac{2}{3}} = {(4)}^{1/3}$