This is not homework, per se, but I have recently started reading Courant and Robbins'(adsbygoogle = window.adsbygoogle || []).push({}); What is Mathematics?

1. The problem statement, all variables and given/known data

Prove by mathematical induction.

(1+q)(1+q^{2})(1+q^{4}) ... (1+q^{2n}) = (1-q^{2n+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 A_{1}was true.

A_{1}= (1+q)(1+q^{2}) = 1+q+q^{2}+q^{3}

Furthermore,

(1-q^{21+1}) / (1-q) = (1-q^{4}) / (1-q) = (1-q^{2})(1+q^{2}) / (1-q) = (1+q)(1+q^{2})

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+q^{2})(1+q^{4}) ... (1+q^{2r}) = (1-q^{2r+1}) / (1-q)

2. (1+q)(1+q^{2})(1+q^{4}) ... (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.

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

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