Explanation

(a)
Let $S = 1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \frac{14}{3^{4}} + ...$
$\Rightarrow S - 1 = \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \frac{14}{3^{4}} = ...$ …(i)
$\Rightarrow \frac{S - 1}{3} = \frac{2}{3^{2}} + \frac{6}{3^{3}} + \frac{10}{3^{4}} + \frac{14}{3^{5}} + \ldots$ …(ii)
On subtracting Eq.(ii) from Eq. (i), we get
$\frac{2}{3}(S - 1) = \frac{2}{3} + \frac{4}{3^{2}} + \frac{4}{3^{3}} + \frac{4}{3^{4}} + ...$
$\Rightarrow S - 1 = 1 + \frac{2}{3} + \frac{2}{3^{2}} + \frac{2}{3^{3}} + ...$
$\Rightarrow \ \ S = 2 + \frac{3}{1 - \frac{1}{3}}$
$= 2 + 1 = 3$