+981
Let \(f(x) = \left\lfloor\dfrac{2 - 3x}{x + 5}\right\rfloor\). Evaluate \(f(1)+f(2) + f(3) + \cdots + f(999)+f(1000).\)
+934 Edit: Wrong Answer
f(1)= -1 and f(1000)= -3. The sum of the terms is equal to the average multiplied by the number of terms, so 1000(-1+-3)/2= -2000.
*Check me if I'm wrong
Hope it helps!
+128460 Look at the graph here...https://www.desmos.com/calculator/nofaug4qzu
And note that
f(1) = -1 f(4) = -2 f(7) = -2 f(10) = -2
f(2) = -1 f(5) = -2 f(8) = -2 f(11) = -2
f(3) = -1 f(6) = -2 f(9) = -2 f(12) = -2
f(13) = -3
...
f(1000) = -3
This implies that the sum is (-1)*3 + (-2)*9 + (1000 - 13 + 1) (-3) = -2985
