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Induction Problem

  • Thread starter dtl42
  • Start date
116
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1. Homework Statement
Prove by Induction that [tex]x^{k}-1=(x-1)(x^{k-1}+x^{k-2}+...+1)[/tex]


2. Homework Equations
None? I need to prove [tex]x^{k+1}-1=(x-1)(x^{k}+x^{k-1}+...+1)[/tex] by assuming this is true: [tex]x^{k}-1=(x-1)(x^{k-1}+x^{k-2}+...+1)[/tex]


3. The Attempt at a Solution
[tex]x^{k+1}-1=(x-1)(x^{k}+x^{k-1}+...+1)[/tex] - I tried substituting in the [tex]x^{k}[/tex] term, and expanding [tex]x*x^{k}-1[/tex], but it failed.
 

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
41,732
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1. Homework Statement
Prove by Induction that [tex]x^{k}-1=(x-1)(x^{k-1}+x^{k-2}+...+1)[/tex]


2. Homework Equations
None? I need to prove [tex]x^{k+1}-1=(x-1)(x^{k}+x^{k-1}+...+1)[/tex] by assuming this is true: [tex]x^{k}-1=(x-1)(x^{k-1}+x^{k-2}+...+1)[/tex]


3. The Attempt at a Solution
[tex]x^{k+1}-1=(x-1)(x^{k}+x^{k-1}+...+1)[/tex]
You can't use that- that's what you are trying to prove.

- I tried substituting in the [tex]x^{k}[/tex] term, and expanding [tex]x*x^{k}-1[/tex], but it failed.
You have xk+1- 1 and you need to use the fact that xk- 1= (x-1)(xk-1+...+ x+ 1). Have you tried dividing xk+1-1 by xk- 1? It does not divide evenly, you get
[tex]x^{k+1}-1= (x^k-1)(x+1)+ \frac{x-1}{x^k-1}[/tex]
but now you should be able to simplify that fraction.
 
116
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That fraction would simplify to [tex]\frac{1}{x^{k}+x^{k-1}+...+1}[/tex] right?
 
116
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That fraction would simplify to [tex]\frac{1}{x^{k}+x^{k-1}+...+1}[/tex] right?
 
116
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Bump, I could use some help
 
Gib Z
Homework Helper
3,344
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Check the power of the leading term in the denominator again.
 
144
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there is another method without any division. When you have [tex]x^{k+1}-1=x \cdot x^k-1[/tex], substitute the x part inte another appropriate term that equals x, then use the induction assumption.
 
Last edited:
116
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What would the other appropriate term be? I can't really work that out in my head, I thought I should substitute for [tex] x^{k} [/tex] because I can separate it more easily.
 
144
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I can give you an example of what technique I mean.
Prove that [tex]3^{2n+1} + 4^{2n+1}[/tex] always is divisible by 7.
Indcution. True for n=0, now assume true for n and consider n+1:
[tex]a_{n+1}=3^{2(n+1)+1} + 4^{2(n+1)+1}=9\cdot 3^{2n+1}+16 \cdot 4^{2n+1}=[/tex]
[tex]=9\cdot 3^{2n+1}+(9+7) \cdot 4^{2n+1}=9 \cdot a_n + 7 \cdot 4^{2n+1}[/tex]
which is divisible by 7 if a_n is.
How can you apply this method on your problem? how can the two terms look like, if they should equal x? :)
 
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