1. Sep 21, 2011

Roscoe1989

1. can someone tell me what r^(n+1)*(1-r)=? I need it to solve a proof, but my math is so rusty! thanks!

2. r^(n+1)*(1-r)

2. Sep 21, 2011

flyingpig

What is the question...? Like the real question.

3. Sep 21, 2011

Roscoe1989

prove by induction on n that
1+r+r^2...+r^=1-r^n+1/1-r
if r doesn't equal 1

4. Sep 21, 2011

flyingpig

Just to be clear, you have

$$r^n + \frac{1}{1 -r}$$ on your right hand side?

5. Sep 21, 2011

Roscoe1989

oopps, sorry. it should be

1+r+...+r^n=1-r^(n+1) / 1-r

6. Sep 21, 2011

flyingpig

7. Sep 21, 2011

flyingpig

Also, shouldn't |r|<1?

8. Sep 21, 2011

Roscoe1989

the what case?

9. Sep 21, 2011

flyingpig

Base Case. You know that step you do in Induction

10. Sep 21, 2011

Roscoe1989

i don't know?

11. Sep 21, 2011

flyingpig

What can n be and cannot be?

12. Sep 21, 2011

flyingpig

edit; mistake

13. Sep 21, 2011

Roscoe1989

is it n can't be 0? but can be anything else?

14. Sep 21, 2011

Dick

If you don't know induction, just multiply out (1+r+...+r^n)*(1-r). Many terms cancel.

15. Sep 21, 2011

Roscoe1989

my textbook is saying 1-r^n+1

16. Sep 21, 2011

flyingpig

Oh sorry I didn't see the 1 in front earlier

Last edited: Sep 21, 2011
17. Sep 21, 2011

Dick

18. Sep 21, 2011

flyingpig

try pluggin in 0

19. Sep 21, 2011

Roscoe1989

n has to be greater than 0

20. Sep 21, 2011

flyingpig

Why?

21. Sep 21, 2011

flyingpig

$$1 + r + r^2 + ... + r^n = \frac{1 - r^{n+1}}{1-r}$$

Is what you have. What happens if n = 0?

22. Sep 21, 2011

Roscoe1989

then the right hand side will equal 1. right?

23. Sep 21, 2011

flyingpig

And what about the left hand side? What happens there?

24. Sep 21, 2011

Dick

Why not? I would interpret 1+r+...+r^n for n=0 to be 1. (1-r^(0+1))/(1-r)=(1-r)/(1-r)=1. I'm not sure you are leading this in a helpful direction, flyingpig.

25. Sep 21, 2011

Roscoe1989

honestly, i don't know