+1858 Let f(x) = \left \lfloor \frac{2 - 3x}{x + 3} + x^2 \right \rfloor.
Find f(1) + f(2) + f(3) + ... + f(999) + f(1000).
+129657 \( \left \lfloor \frac{2 - 3x}{x + 3} + x^2 \right \rfloor \)
f(1) = 0
f(2) = 3
f(3) = 7
f(4) = 14
f(5) = 23
f(6) = 34
f(7) = 47
f(8) = 62
f(9) = 78
.
.
.
f (1000) = 999,997
It appears that, from x = 3 to x = 8 inc;usive, f(x) = x^2 - 2
And from x = 9 to x =1000 inclusive , f(x) = x^2 - 3
So
So....the sum is
8 1000
3 + Σ x^2 - 2 + Σ x^2 - 3 = 333,830,510
x = 3 x = 9
