# Homework Help: Mathematical Induction: Chapter 1, Example 4 from WiM? by Courant

1. Jan 8, 2010

### DarrenM

This is not homework, per se, but I have recently started reading Courant and Robbins' What is Mathematics?

1. The problem statement, all variables and given/known data
Prove by mathematical induction.
(1+q)(1+q2)(1+q4) ... (1+q2n) = (1-q2n+1) / (1-q)

2. Relevant equations
Only the problem itself.

3. The attempt at a solution
Using the method the book briefly introduced, I first established that A1 was true.

A1 = (1+q)(1+q2) = 1+q+q2+q3

Furthermore,

(1-q21+1) / (1-q) = (1-q4) / (1-q) = (1-q2)(1+q2) / (1-q) = (1+q)(1+q2)

Next, I attempted to show that Ar+1 follows from Ar, but this is where I ran into trouble.

1. (1+q)(1+q2)(1+q4) ... (1+q2r) = (1-q2r+1) / (1-q)

2. (1+q)(1+q2)(1+q4) ... (1+q2r)(1+q2r+1) = {(1-q2r+1) / (1-q)}(1+q2r+1)

This is the point at which I hit a wall. I feel like it's something absurdly simple, but I've been stuck on this for hours; I think it's the variable raised to a power raised to a binomial power that is throwing me for a loop.

I tried simply multiplying (1-q2r+1) and (1+q2r+1); that gives me another binomial that is the difference of two squares:

{(1-(q2r+1)(q2r+1)} / (1-q)

But again, stumped. I believe that if I multiply those two terms together I get (1-q2r+1+2r+1) / (1-q). If that's the case, this is the point at which I have completely lost it. However, I am not confident that I'm doing that properly.

Just to see if I could work backwards, I took a look at what I think Ar+1 would look like:

Ar+1 = (1-q2r+2) / (1-q)

Any hints on where/what I'm doing wrong would be very much appreciated.

P.S. - All of this is very new to me, so I apologize if it's not quite as clear or as polished as it should be.

2. Jan 8, 2010

### icystrike

well..
$$A_{r+1}=(1+q^{2^{r+1}})\frac{1-q^{2^{r+1}}}{1-q}$$
Just by focusing on the numerator,
$$(1+q^{2^{r+1}})(1-q^{2^{r+1}})$$
Gives us
$$1-q^{2^{r+1}\times2}$$

I'll leave the rest to you.
Whenever you are stuck, check your previous workings , then try other approach to that question.

3. Jan 8, 2010

### DarrenM

Sorry about the parenthesis and brackets, I'm not sure how to clearly represent fractions with the forum-code.

And I'm afraid that 1-q2r+1x2 takes me right back to where I was stuck. Any additional hint to nudge me in the right direction? :)

4. Jan 8, 2010

### icystrike

law of indices:

$$a^b \times a^c$$=$$a^{b+c}$$

Now:
$$2^{r+1} \times 2^{1}$$ = ?

Last edited: Jan 8, 2010
5. Jan 8, 2010

### DarrenM

Aaaaargh.

It's not like I didn't just do that this morning on a previous problem. Nor is it like I wasn't staring at 1-q2(2r+1) while trying to solve this.

2r+1 x 21 = 2r+1+1 = 2r+2

Thanks very much for the help.

6. Jan 8, 2010

### icystrike

Welcome! Never ever abandon your law of indices agn (=