Why Is My Algebra Proof Not Working?

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Greetings,

It's been awhile since I've done induction or proofs in general, but I could not figure out where I went wrong on this one for the life of me. If anyone has an idea it would be much appreciated. I've uploaded a picture of the problem in the book as well as of my work. I thought I did everything correct but the algebra isn't working out.

Thanks all.

Problem (12): http://imageshack.us/m/30/7931/20110512190557593.jpg

Attempted Solution: http://imageshack.us/m/812/3723/2011051222200544.jpg
 
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What is 2*(2^n) equal to? [Hint: not 2^(2n)]
 
Right, so isn't 2*(2^n) = 2^(n+1) ?

But then I end up with 2^(2n) = 2^(n+1) which eventually leads to n = 1, yes?
 
EsponV said:
Right, so isn't 2*(2^n) = 2^(n+1) ?

But then I end up with 2^(2n) = 2^(n+1) which eventually leads to n = 1, yes?

I don't understand how you could end up with 2^{2n}=2^{n+1} again. If you apply the correct formula in the line

(2^n-1)+2^{(n+1)-1}=2^{n+1}-1

then surely you'd end up with something else??
 
Ah, I had been incorrectly stating that 2^n + 2^n = 2^2n instead of 2*2^n = 2^n+1. Thanks for the help, sorry about the algebra error.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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