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Let

\( \[f(x) = \left\lfloor\frac{2 - 3x}{x + 3}\right\rfloor.\]\)
Evaluate  \( $f(1)+f(2) + f(3) + \dots + f(999)+f(1000).$\)

 Oct 22, 2023
 #1
avatar+691 
-1

We can find the first few terms of the sum to see if there is a pattern:

f(1) = -1 f(2) = -1 f(3) = -1 f(4) = -2 f(5) = -2 f(6) = -2 f(7) = -3 f(8) = -3 f(9) = -3

We see that the function values repeat in a cycle of length 3: −1, −1, −2, −2, −2, −3, −3, −3. Since there are 1000 terms in the sum, we can divide 1000 by 3 to find that there are 333 complete cycles of length 3, plus one additional term. The sum of the terms in each complete cycle is −1−1−2=−4, so the sum of the first 333 cycles is −4⋅333=−1332. The additional term is f(1000)=−3, so the total sum is −1332−3=∗∗−1335∗∗.

Therefore, the answer is -1335.

 Oct 27, 2023

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