What is the solution to r^(n+1)*(1-r) in calculus?

[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)

 

[flyingpig]

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

 

[Roscoe1989]

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

 

[flyingpig]

Just to be clear, you have

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

 

[Roscoe1989]

oopps, sorry. it should be

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

 

[flyingpig]

What is the base case? Let's start with that

 

[flyingpig]

Also, shouldn't |r|<1?

 

[Roscoe1989]

the what case?

 

[flyingpig]

Base Case. You know that step you do in Induction

 

[Roscoe1989]

i don't know?

 

[flyingpig]

What can n be and cannot be?

 

[flyingpig]

edit; mistake

 

[Roscoe1989]

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

 

[Dick]

oopps, sorry. it should be

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

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

 

[Roscoe1989]

my textbook is saying 1-r^n+1

 

[flyingpig]

my textbook is saying 1-r^n+1

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

 

[Dick]

my textbook is saying 1-r^n+1

Your textbook is right.

 

[flyingpig]

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

try pluggin in 0

 

[Roscoe1989]

n has to be greater than 0

 

[flyingpig]

n has to be greater than 0

Why?

 

[flyingpig]

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

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

 

[Roscoe1989]

then the right hand side will equal 1. right?

 

[flyingpig]

then the right hand side will equal 1. right?

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

 

[Dick]

Why?

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.

 

[Roscoe1989]

honestly, i don't know

 

[flyingpig]

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.

Nononon, Ros asked if it can be 0 before

 

[Roscoe1989]

does it mean that it's "proven"?

 

[flyingpig]

honestly, i don't know

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

You were originally given this. You got it right when you said the rhs is 1 when n = 0

 

[flyingpig]

does it mean that it's "proven"?

By Induction? No

If only though, if only...

 

[Roscoe1989]

but what does it mean on the LHS?