Let f(x) = lfloor (2 - 3x)/(3x + 8) \rfloor Evaluate $f(1)+f(2) + f(3) + \dots + f(999)+f(1000).$ (This sum has $1000$ terms, one for the result when we input each integer from $1$ to $1000$ into $f$.)
floor [ (2 -3x) / (3x + 8) ]
The value of f(1) to f(1000) always = -1
So....the sum is -1000
+677 
