Explanation

(b)
Given, $P(X = k) = \frac{(k + 1)a}{3^{k}},$ for $x \in \{ 0,1,2,\ldots\infty\}$
As we know that
$P(0) + P(1) + P(2) + ...\infty = 1$
$\Rightarrow \ a + \frac{2a}{3} + \frac{3a}{3^{2}} + ...\infty = 1$ ...(i)
Let $S = a\left( 1 + \frac{2}{3} + \frac{3}{3^{2}} + \frac{4}{3^{3}} + ...\infty \right)$
$\Rightarrow \ \frac{\frac{1}{3}S = a\left( \frac{1}{3} + \frac{2}{3^{2}} + \frac{3}{3^{3}} + ...\infty \right)}{S - \frac{1}{3}S = a\left( 1 + \frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + ...\infty \right)}$
$\Rightarrow \ \frac{2}{3}S = a\left( \frac{1}{1 - \frac{1}{3}} \right)\ \Rightarrow \ S = \frac{9a}{4}$
$\therefore$ From Eq.(i)
$\frac{9a}{4} = 1\ \Rightarrow \ \ a = \frac{4}{9}$