Question

PLEASE HELP

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.)

Answer 1

The sum
`(f(1) + f(2) + \ldots + f(999) + f(1000)\)` evaluates to -6.

Explanation:

Let's break down the expression for ( f(x) ).

`\[ f(x) = \lfloor 2 - (3x)/(x + 3) \rfloor \]`

To simplify the expression inside the floor function, we first combine the terms:

`\[ f(x) = \lfloor (2x + 6 - 3x)/(x + 3) \rfloor \]`

`\[ f(x) = \lfloor (6 - x)/(x + 3) \rfloor \]`

Now, let's analyze the expression
`\( (6 - x)/(x + 3) \)`:

1. If ( x < 6 ), then
`\( (6 - x)/(x + 3) > 0 \)`, and
`\( \lfloor (6 - x)/(x + 3) \rfloor = 0 \).`

2. If ( x = 6 ), then
`\( (6 - x)/(x + 3) = 0 \)`, and
`\( \lfloor (6 - x)/(x + 3) \rfloor = 0 \).`

3. If
`\( x > 6 \)`, then
`\( (6 - x)/(x + 3) < 0 \)`, and
`\( \lfloor (6 - x)/(x + 3) \rfloor = -1 \).`

Now, consider the values of ( x ) in the given range (1 to 1000):

1. For ( x = 1 ) to ( x = 6 ), ( f(x) = 0 ).

2. For ( x = 7 ) to ( x = 1000 ), ( f(x) = -1 ).

So, the sum \( f(1) + f(2) + \ldots + f(999) + f(1000) \) is equivalent to the sum of 994 zeros and 6 -1's:

`\[ 994 * 0 + 6 * (-1) = 0 - 6 = -6 \]`

Therefore, the value of the given sum is -6.