Explanation

(d)
Let $1 + \frac{1}{3}x + \frac{1 \bullet 4}{3 \bullet 6}x^{2} + \frac{1 \bullet 4 \bullet 7}{3 \bullet 6 \bullet 9}x^{3} + ... = {(1 + y)}^{n}$
$= 1 + ny + \frac{n(n - 1)}{2!}y^{2} + ...$
On comparing the terms, we get
$ny = \frac{1}{3}x,\ \frac{n(n - 1)}{2!}\ y^{2} = \frac{1 \bullet 4}{3 \bullet 6}x^{2}$
On solving, we get
$n = - \frac{1}{3},\ \ \ \ \ \ \ y = - x$
$\therefore$ Required expansion is ${(1 - x)}^{\frac{- 1}{3}}$