Sequences full of fractions

Relevant wiki: Recurrence Relations - Method of Summation Factors

Getting rid of denominators, we obtain $(n + 4)a_n = (n + 3)a_{n-1} + (n + 2)(n + 1)$ Let $b_n = (n + 4)a_n$ . Then $b_{n-1} = (n + 3)a_{n-1}$ , and the recurrence relation becomes $b_n = b_{n - 1} + (n^2 + 3n + 2)$ Furthermore, $b_0 = (0 + 4)a_0 = 4$ . Therefore, by definition, the closed form of $b_n$ is $b_n = 4 + \sum_{k = 1}^n {(k^2 + 3k + 2)} = 4 + \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} + 2n$ Since $b_n = (n + 4)a_n$ , we now have $a_n = \frac{1}{n+4}\left( 4 + \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} + 2n \right)$ and $a_{40} = \frac{1}{40+4}\left( 4 + \frac{40(40+1)(2(40)+1)}{6} + \frac{3(40)(40+1)}{2} + 2(40) \right) = \boxed{561}$