Explanation

(a)
$\left( 1 - 2x - x^{2} \right)(e^{x})$
$= (1 - 2x - x^{2})\left( 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... + \frac{x^{k}}{k!} + ...\infty \right)$
$= \left( 1 + x + \frac{x^{2}}{2!} + ... + \frac{x^{k}}{k!} + ...\infty \right)$
$- 2\left( x + x^{2} + \frac{x^{3}}{2!} + ... + \frac{x^{k}}{(k - 1)!} + \frac{x^{k + 1}}{k!} + ...\infty \right)$
$- \left( x^{2} + x^{3} + \frac{x^{4}}{2!} + \ldots\frac{x^{k}}{(k - 2)!} + \frac{x^{k + 1}}{(k - 1)!} + \frac{x^{k + 2}}{k!} + ...\infty \right)$
$\therefore$Cofficient of$\ x^{k}$in $\left( \frac{1 - 2x - x^{2}}{e^{- x}} \right) = \frac{1}{k!} - \frac{2}{(k - 1)!} - \frac{1}{(k - 2)!}$
$= \frac{1}{k!} - \frac{2k}{k!} - \frac{k(k - 1)}{k!}$
$= \frac{1 - k - k^{2}}{k!}$