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Let
f(x) = \left\lfloor\frac{2 - 3x}{3x + 1}\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.)

 Jan 13, 2024
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\(f(x) = \left\lfloor\frac{2 - 3x}{3x + 1}\right\rfloor\)

When x = 1, f(1) = -0.25 rounded down = -1

When x = 2, f(2) = -4/7 rounded down = -1

When x = 3, f(3) = -0.7 rounded down = -1

Let's try bigger values of x to see if the answer is less than -1.

When x = 33, f(33) = -0.97 rounded down = -1.

 

For any x value, the answer rounded down will be -1, so we get:

 

-1 -1 -1 -1 -1 - ... -1 -1 = -1000

 

Answer: -1000

 Jan 13, 2024

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