+0

0
5
3
+68

Let

$$f(x) = \left\lfloor\frac{2 - 3x}{x + 3}\right\rfloor.$$
Evaluate  $$f(1)+f(2) + f(3) + \dots + f(999)+f(1000).$$(This sum has  terms, one for the result when we input each integer from  to  into .)

Mar 24, 2024

#3
+128845
+1

f(1)   = -1

f(2)  =  -1

f(3)  = -2

.

.

.

f(8)  = -2

f(9) =  -3

.

.

.

f(1000) = -3

Sum of terms

2(-1)  + 6(-2) + 992(-3)   =    -2990

Mar 24, 2024

#1
+179
+1

Analyzing the function f(x) = floor((2 - 3x) / (x + 3)):

The denominator (x + 3) is 0 when x = -3. This means f(x) is undefined at x = -3.

We will need to consider different cases based on the sign of the denominator (x + 3) and the relative values of 2 - 3x compared to 0.

Cases for f(x):

x < -3: In this case, both denominator (x + 3) and numerator (2 - 3x) are negative. Dividing two negative numbers results in a positive value.

Since we take the floor (greatest integer less than or equal to), f(x) will be -1.

-3 < x < 2/3: Here, the denominator (x + 3) is positive, but the numerator (2 - 3x) is negative.

Dividing a positive by a negative results in a negative number.

Taking the floor of a negative number keeps it negative, so f(x) will be -2.

x = 2/3: At this specific point, the numerator becomes 0, and the result of the division is 0. The floor of 0 is 0, so f(x) = 0.

x > 2/3: In this case, both the numerator (2 - 3x) and denominator (x + 3) are positive. Dividing two positive numbers results in a positive value.

Taking the floor doesn't change the positive sign, so f(x) will be 1.

Evaluating the sum:

The key to evaluating the sum efficiently is to recognize that for a large range of x values (between -3 and 2/3), f(x) will be -2.

We can exploit this by calculating the number of terms that fall into this range and summing the contributions from the remaining terms separately.

Number of terms where f(x) = -2:

We know x = -3 falls outside this range (f(x) is undefined).

The range ends when x = 2/3, which is between terms 1000 and 1001 (1000th term is x = 999 and 1001st term is x = 1000).

Therefore, there are 1000 - (-3) + 1 = 1004 terms where f(x) = -2.

Contribution from terms where f(x) = -2:

Each term contributes -2 to the sum.

Total contribution = -2 * (number of terms) = -2 * 1004 = -2008

Remaining terms:

We need to consider terms for x < -3 (f(x) = -1), x = 2/3 (f(x) = 0), and x > 2/3 (f(x) = 1).

There are very few terms less than -3 (all negative x values), and they can be ignored for a large sum like this (their contribution will be negligible).

There's only one term for x = 2/3, contributing f(2/3) = 0.

The remaining terms from x slightly greater than 2/3 to x = 1000 will all have f(x) = 1. The exact number of these terms depends on the specific values, but there will be significantly fewer compared to the 1004 terms with f(x) = -2.

Overall Sum:

Sum from terms with f(x) = -2: -2008

Contribution from f(2/3) (x = 2/3): 0

Contribution from remaining terms with f(x) = 1 (positive but less than those with -2): + (positive value)

Since the number of terms with f(x) = 1 is significantly less than those with -2, and there's a negligible contribution from terms less than -3, the positive value from the remaining terms will be much smaller than 2008.

Therefore, the overall sum f(1) + f(2) + ... + f(999) + f(1000) is equal to -2008.

Mar 24, 2024
#2
+1618
+1

To evaluate the expression f(1) + f(2) + f(3) + ... + f(999) + f(1000), we need to understand the behavior of the function f(x) first.

The function f(x) is defined as the floor of (2 - 3x)/(x + 3). The floor function rounds down to the nearest integer. We can observe the behavior of this function for different values of x to understand how to proceed.

Let's analyze the function for a few values of x:

f(1) = floor((-1)/4) = -1

f(2) = floor((-4)/5) = -1

f(3) = floor((-7)/6) = -2

f(4) = floor((-10)/7) = -2

f(5) = floor((-13)/8) = -2

It seems that f(x) remains constant within intervals of x before decreasing by 1 and remaining constant again.

Now let's find the length of each interval:

For x = 1 to x = 4, f(x) = -1.

For x = 5 to x = 8, f(x) = -2.

This pattern continues. So, for each interval of length 4, the value of f(x) remains constant, then decreases by 1.

The sum of all f(x) from x = 1 to x = 1000 can be calculated by dividing 1000 by 4 to find out how many complete cycles occur, and then multiplying by the sum of each cycle plus the remaining values.

Number of complete cycles: 1000/4 = 250.

In each cycle:

For x = 1 to x = 4, the sum is -1 -1 -1 -1 = -4.

For x = 5 to x = 8, the sum is -2 -2 -2 -2 = -8.

So, the sum of each complete cycle is -4 -8 = -12.

The remaining values are f(997), f(998), f(999), f(1000). These are -3, -3, -3, -3 respectively.

So, the total sum is 250 * (-12) - 12 = -3000 - 12 = -3012.

Mar 24, 2024
#3
+128845
+1

f(1)   = -1

f(2)  =  -1

f(3)  = -2

.

.

.

f(8)  = -2

f(9) =  -3

.

.

.

f(1000) = -3

Sum of terms

2(-1)  + 6(-2) + 992(-3)   =    -2990

CPhill Mar 24, 2024