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Let

f(x) = floor(2 - 3x/x + 3).
Evaluate f(1)+f(2) + f(3) + ... + f(999)+f(1000). (This sum has 1000 terms, one for the result when we input each integer from 1 to 1000 into f.)

User Chaldaean
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Answer:

To evaluate the sum f(1) + f(2) + f(3) + ... + f(999) + f(1000), we need to substitute each integer from 1 to 1000 into the function f(x) = floor(2 - 3x/x + 3) and then add up the results.

User Kazuhiko Nakayama
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