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

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

The answer is 2026.

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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