Explanation

(c)
Let $S = \sum_{n = 1}^{\infty}\frac{2n^{2} + n + 1}{n!}$
$= \sum_{n = 1}^{\infty}\left( \frac{2n}{(n - 1)!} + \frac{1}{(n - 1)!} + \frac{1}{n!} \right)$
$= \sum_{n = 1}^{\infty}\left( \frac{2}{(n - 2)!} + \frac{3}{(n - 1)!} + \frac{1}{n!} \right)$
$= 2\left( 1 + \frac{1}{1!} + \frac{1}{2!} + ...\infty \right) + 3\left( 1 + \frac{1}{1!} + ...\infty \right) + \left( \frac{1}{1!} + \frac{1}{2!} + ... \right)$
$= 2e + 3e + e - 1$
$= 6e - 1$