Explanation

(d)
Given, $\frac{1}{e^{3x}}\left( e^{x} + e^{5x} \right) = a_{0} + a_{1}x + a_{2}x^{2} + ...$
$\Rightarrow \left( e^{- 2x} + e^{2x} \right) = a_{0} + a_{1}x + a_{2}x^{2} + ...$
$\Rightarrow 2\left\lbrack 1 + \frac{(2x)^{2}}{2!} + \frac{(2x)^{4}}{4!} + ... \right\rbrack = a_{0} + a_{1}x + a_{2}x^{2} + ...$
$\Rightarrow a_{1} = a_{3} = a_{5} = ... = 0$
$\therefore 2a_{1} + 2^{3}a_{3} + 2^{5}a_{5} + ... = 0$