Explanation

(c)
$\frac{e^{7x} + e^{3x}}{e^{5x}} = e^{2x} + e^{- 2x}$
$= \left\lbrack 1 + \frac{2x}{1!} + \frac{(2x)^{2}}{2!} + \frac{(2x)^{3}}{3!} + ...\infty \right\rbrack$
$+ \left\lbrack 1 - \frac{2x}{1!} + \frac{(2x)^{2}}{2!} - \frac{(2x)^{3}}{3!} + ...\infty \right\rbrack$
$\Rightarrow e^{2x} + e^{- 2x} = 2\left\lbrack 1 + \frac{(2x)^{2}}{2!} + \frac{(2x)^{4}}{4!} + ...\infty \right\rbrack$
Hence, the constant term is 2.