# Can someone double check this, or tell me happened?

1. Apr 27, 2004

### JasonRox

I don't understand how they got rid of the negative.
l = absolute, but they aren't used anyways.

$$\mid \S_n - frac{a}{1 - R} \mid = \mid a( \frac{1 - r^n}{1-r} ) - \frac{a}{1-r}\mid$$

$$= \mid \frac{a r^n}{1-r} \mid$$

I'm getting...

$$= \mid \frac{-a r^n}{1-r} \mid$$

I always miss negatives, but I don't think I did this time.

Last edited: Apr 28, 2004
2. Apr 27, 2004

The absolute value's going to kill the negative either way.

3. Apr 28, 2004

### krab

Did you know that $|ab|=|a||b|$? Therefore $|-b|=|-1||b|=|b|$.

4. Apr 28, 2004

### JasonRox

The negative is gone after we use the absolutes.

We haven't used them, so where did it go?

5. Apr 28, 2004

### philosophking

I see what you did wrong. When you factor out the _a_ at the beginning, you get Sn -1, which is r^n -1, and that should solve your problem.

6. Apr 28, 2004

### JasonRox

Sn isn't even involved.

7. Apr 28, 2004

### JasonRox

I'll break it down even more, and I hope I find my mistake.

$$\mid S_n - \frac{a}{1 - R} \mid = \mid a( \frac{1 - r^n}{1-r} ) - \frac{a}{1-r}\mid$$

$$= \mid \frac{a - ar^n}{1 - r} - \frac{a}{1 - r} \mid$$

...same denominator...

$$= \mid \frac{a - ar^n - a}{1 - r} \mid$$

$$= \mid \frac{-ar^n}{1 - r} \mid$$

8. Apr 28, 2004

### HallsofIvy

Staff Emeritus
And you said before
What do you mean "we haven't used them"? You certainly are using the absolute value. That's why it's there and that's why
$$= \mid \frac{-ar^n}{1 - r} \mid = \frac{ar^n}{1-r}$$

If you mean that there is still an absolute value sign in the answer, that doesn't mean they haven't been used ||a||= |a| (absolute value of the absolute value of a is the same as the absolute value of a).

9. Apr 30, 2004

### JasonRox

Nevermind.

I shouldn't have put the absolutes in to begin with. It changes nothing!

It's like we right the number 4 instead of 4^1, which is the way it should be done.

There is no reason for the negative to "disappear" on its own.

Yes, I understand absolutes, but they only work if you use them.

10. Apr 30, 2004