16.6

Algebra Level 4

$\large \sum_{n=0}^{2016} \left \lfloor \frac{(n+3)(n+4)}{(n+1)(n+2)} \right \rfloor = \, ?$

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

1 solution

Matteo Monzali
Jun 13, 2016

From 0 to 4 the values of the expression are 6, 3, 2 and 2; from that point the values are always 1, so we can do (2016-4+1)+6+3+2+2=2026