Let
\[f(x) = \left\lfloor\frac{3x - 5}{x + 4}\right\rfloor.\]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 ${}f$.)

To evaluate the sum $ f(1) + f(2) + f(3) + \dots + f(999) + f(1000) $ where

\[
f(x) = \left\lfloor \frac{3x - 5}{x + 4} \right\rfloor,
\]

we start by simplifying the expression inside the floor function:

\[
\frac{3x - 5}{x + 4}.
\]

Let's perform polynomial long division to express this fraction as a sum of a linear term and a fractional term:

1. Divide $ 3x $ by $ x $ to get 3.
2. Multiply $ 3 $ by $ x + 4 $ to get $ 3x + 12 $.
3. Subtract $ 3x + 12 $ from $ 3x - 5 $:

\[
3x - 5 - (3x + 12) = -17.
\]

Thus,

\[
\frac{3x - 5}{x + 4} = 3 - \frac{17}{x + 4}.
\]

So,

\[
f(x) = \left\lfloor 3 - \frac{17}{x + 4} \right\rfloor.
\]

To find the value of $ f(x) $, notice that $ 3 - \frac{17}{x + 4} $ is a number slightly less than 3, because $ \frac{17}{x + 4} $ is positive. Therefore,

\[
f(x) = \left\lfloor 3 - \frac{17}{x + 4} \right\rfloor = 2.
\]

$ f(x) = 2 $ holds as long as

\[
3 - \frac{17}{x + 4} \geq 2,
\]

which simplifies to:

\[
-\frac{17}{x + 4} \geq -1 \quad \text{or} \quad \frac{17}{x + 4} \leq 1.
\]

Cross-multiplying gives:

\[
17 \leq x + 4 \quad \text{or} \quad x \geq 13.
\]

Thus, for $ x \geq 13 $, $ f(x) = 2 $.

Next, let's check the values of $ f(x) $ for $ x < 13 $:

- For $ x = 1 $, $ f(1) = \left\lfloor \frac{-2}{5} \right\rfloor = -1 $.
- For $ x = 2 $, $ f(2) = \left\lfloor \frac{1}{6} \right\rfloor = 0 $.
- For $ x = 3 $, $ f(3) = \left\lfloor \frac{4}{7} \right\rfloor = 0 $.
- For $ x = 4 $, $ f(4) = \left\lfloor \frac{7}{8} \right\rfloor = 0 $.
- For $ x = 5 $, $ f(5) = \left\lfloor \frac{10}{9} \right\rfloor = 1 $.
- For $ x = 6 $, $ f(6) = \left\lfloor \frac{13}{10} \right\rfloor = 1 $.
- For $ x = 7 $, $ f(7) = \left\lfloor \frac{16}{11} \right\rfloor = 1 $.
- For $ x = 8 $, $ f(8) = \left\lfloor \frac{19}{12} \right\rfloor = 1 $.
- For $ x = 9 $, $ f(9) = \left\lfloor \frac{22}{13} \right\rfloor = 1 $.
- For $ x = 10 $, $ f(10) = \left\lfloor \frac{25}{14} \right\rfloor = 1 $.
- For $ x = 11 $, $ f(11) = \left\lfloor \frac{28}{15} \right\rfloor = 1 $.
- For $ x = 12 $, $ f(12) = \left\lfloor \frac{31}{16} \right\rfloor = 1 $.

The results are:

- $ f(1) = -1 $
- $ f(2) = f(3) = f(4) = 0 $
- $ f(5) = f(6) = f(7) = f(8) = f(9) = f(10) = f(11) = f(12) = 1 $
- $ f(x) = 2 $ for $ x \geq 13 $.

The sum $ f(1) + f(2) + \dots + f(1000) $ can now be computed by adding:

- The first term, $ f(1) = -1 $.
- The next three terms, $ f(2) + f(3) + f(4) = 0 $.
- The next eight terms, $ f(5) + \dots + f(12) = 8 $.
- The remaining $ 988 $ terms (from $ f(13) $ to $ f(1000) $), each equal to 2: $ 2 \times 988 = 1976 $.

Adding these gives:

\[
\text{Total sum} = -1 + 0 + 8 + 1976 = 1983.
\]

Thus, the final answer is:

\[
\boxed{1983}.
\] 

Would you like more details or have any questions? Here are some related questions you might find useful:

1. What is the general process for polynomial long division?
2. How do floor functions behave with fractional inputs?
3. Why does $ f(x) $ change at $ x = 13 $ in this problem?
4. What other types of problems involve floor functions?
5. How can we graph the function $ f(x) $ in this problem?

**Tip**: Always double-check the boundary values in problems involving piecewise functions, as these can often reveal key changes in behavior.

Let f(x) = ⌊(3x - 5) / (x + 4)⌋. 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.)

Discover how to evaluate the sum f(1) + f(2) + ... + f(1000) where f(x) = ⌊(3x - 5) / (x + 4)⌋. This problem involves key algebraic concepts such as the floor function, polynomial division, and integer evaluation, suitable for high school students in Grades 9-12.

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