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

[DarrenM]

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

Homework Statement

Prove by mathematical induction.
(1+q)(1+q²)(1+q⁴) ... (1+q²ⁿ) = (1-q²ⁿ⁺¹) / (1-q)

Homework Equations

Only the problem itself.

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 A_r+1 follows from A_r, but this is where I ran into trouble.

1. (1+q)(1+q²)(1+q⁴) ... (1+q^2r) = (1-q^2r+1) / (1-q)

2. (1+q)(1+q²)(1+q⁴) ... (1+q^2r)(1+q^2r+1) = {(1-q^2r+1) / (1-q)}(1+q^2r+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-q^2r+1) and (1+q^2r+1); that gives me another binomial that is the difference of two squares:

{(1-(q^2r+1)(q^2r+1)} / (1-q)

But again, stumped. I believe that if I multiply those two terms together I get (1-q^2r+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 A_r+1 would look like:

A_r+1 = (1-q^2r+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.

 

[icystrike]

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

I don't understand your parenthesis.
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.

 

[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-q^2r+1x2 takes me right back to where I was stuck. Any additional hint to nudge me in the right direction? :)

 

[icystrike]

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-q^2r+1x2 takes me right back to where I was stuck. Any additional hint to nudge me in the right direction? :)

law of indices:

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

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

 

[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-q^2(2r+1) while trying to solve this.

2^r+1 x 2¹ = 2^r+1+1 = 2^r+2

Thanks very much for the help.

 

[icystrike]

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