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

[Roscoe1989]

ok...so when the right side equals the left side... what does that mean?

 

[Dick]

Yeah I probably wouldn't do that [induction] either. But Ros asked for



in post 3

Ok, missed the "induction" bit in there.

 

[flyingpig]

Dick and flyingpig, you are missing, I think, that roscoe1989 doesn't know what happens in the LHS when n=0.
When n=0, you don't get 1+r²+r³+...+r⁰.
1+r²+r³+...+r^n is just a notation. When n=0 you simply get r⁰ and this is worth 1. So when n=0, both the LHS and RHS is worth 1, so the equality holds, meaning that it does makes sense to consider n=0 (it's a possible/allowed value for n).

Yeah and I am trying to lead Ros to that.

 

[Roscoe1989]

ooooooh, ok

 

[flyingpig]

ok...so when the right side equals the left side... what does that mean?

That means the equal sign worked!

It's probably that I tell you what Induction is (in brief form) first.

(1) Base Case. Usually we take n = 1 or 0. We show that it (S(n)) is true first

(2) Inductive Step. Assume that (1) is true and then you will write your Inductive hypothesis and show that it is true for S(n+1).

 

[flyingpig]

ooooooh, ok

I just led to the base case, so we just did (1).

 

[Roscoe1989]

so then to prove (2), we would have to do n+1, which, in our case, is 0+1=1 and that would give us 1-r^2 / 1-r right?

 

[flyingpig]

Where did you get the 0 from...? I have to go to sleep soon...

 

[Roscoe1989]

sorry, i meant 1+1=2. you've been really helpful, and don't let me keep you awake! :)

 

[flyingpig]

I have to leave now because by the time I get back, I would be very late already. I will leave you with this.

When I said "n + 1" , I mean for a general n. The base case is when we take the most fundamental n. It just happened to be 0.

So you have to show this

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

 

[Roscoe1989]

thanks flyingpig, you've been a great help!