Let

f(x) = \left\lfloor\frac{2 - 3x}{3x + 1}\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.)

eramsby1O1O Jun 3, 2024

#1**+1 **

This problem is actually really really easy!

First off, let's look at the first few terms. \(f(x) = \left\lfloor\frac{2 - 3x}{3x + 1}\right\rfloor \)

f(1) = -1, f(2) = -1, f(3) = -1, f(x) = -1, even f(1000)=-1.

Why is this? This is because the floor symbol means round down no matter what. Since \(2-3x<3x+1\) for \(x>1/6\).

This means we will always have a fraction less than 0 and greater than -1,, meaning it always rounds down to -1. I hope this is a clear explenation.

In the end, we just have \((-1)(1000)=-1000\)

This might be wrong, but I'm not sure.

Thanks! :)

NotThatSmart Jun 3, 2024

#1**+1 **

Best Answer

This problem is actually really really easy!

First off, let's look at the first few terms. \(f(x) = \left\lfloor\frac{2 - 3x}{3x + 1}\right\rfloor \)

f(1) = -1, f(2) = -1, f(3) = -1, f(x) = -1, even f(1000)=-1.

Why is this? This is because the floor symbol means round down no matter what. Since \(2-3x<3x+1\) for \(x>1/6\).

This means we will always have a fraction less than 0 and greater than -1,, meaning it always rounds down to -1. I hope this is a clear explenation.

In the end, we just have \((-1)(1000)=-1000\)

This might be wrong, but I'm not sure.

Thanks! :)

NotThatSmart Jun 3, 2024