Let
\[f(x) = \left\lfloor\frac{2 - 3x}{2x + 8}\right\rfloor.\]
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 ${}f$.)
+129881 Note
f(1) = floor [ -1/10 ] = -1
f(2) = floor[ -4/12 ] = -1
f(3) = floor [ -7 / 14 ] = -1
f(4) = floor [ -10 / 16 ] = -1
.
.
f(10) = floor [-28 /28] = -1
f(11) = floor [ -31 / 30] = -2
.
.
f(1000) = floor [ -2998 / 2008 ] = - 2
We have that the first 10 terms evaluate to -1 and the last 990 evaluate to -2
Sum = 10 (-1) + 990 ( -2) = -1990
