DarrenM
Jan8-10, 08:50 PM
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.
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.