16.15

$\displaystyle \sum_{n = 0}^{2016} \left \lfloor \frac{n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)}{n^9 +1} \right \rfloor =$ $\displaystyle \sum_{n = 0}^{3} \left \lfloor \frac{n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)}{n^9 +1} \right \rfloor =$ $0 + \lfloor \frac{6!}{2} \rfloor + \lfloor \frac{7!}{513} \rfloor + \lfloor \frac{8!}{2\cdot19684} \rfloor = 0 + 360 + 9 + 1 = \boxed{370}$ Note.- Let $p:\mathbb{N} \to \mathbb{R}$ such that $p(n) = \frac{n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)}{n^9 +1}, \space \forall n \in \mathbb{N} .$ Dividing numerator and denominator by $n^6, \space \forall n \ge 1$ we'll get $p(n) = \frac{1(1 + 1/n)(1 + 2/n)(1 + 3/n)(1 + 4/n)(1 + 5/n)}{n^3 +1/n^6}$ the numerator of $p(n)$ is a strictly decreasing function $\forall n \ge 4$ and the denominator of $p(n)$ is a strictly increasing function $\forall n \ge 4$ and this implies that the function $p(n)$ is a strictly decreasing function $\forall n \ge 4$ and $p(4) < \frac{2\cdot2 \cdot2\cdot2 \cdot (2 + 1/4)}{4^3} = \frac{36}{64} < 1$ ...