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prove by induction that ((x^(k))1)=(x1)(x^(k1)+x^(k2)+...+x+1)
i got stuck when i have to prove k=n+1 from k=n being assumed true
 one year ago
 one year ago
prove by induction that ((x^(k))1)=(x1)(x^(k1)+x^(k2)+...+x+1) i got stuck when i have to prove k=n+1 from k=n being assumed true
 one year ago
 one year ago

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satellite73Best ResponseYou've already chosen the best response.0
something is wrong with that formula
 one year ago

johnny0929Best ResponseYou've already chosen the best response.0
what is wrong with the formula?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Looks correct to me.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Anyways, you have to start with a base case of k=1. Then, you have\[(x1)\overset{\text{?}}{=} (x1)(1)\]Which is true, so the base case is good.
 one year ago

johnny0929Best ResponseYou've already chosen the best response.0
yup i have completed the base case and i got stuck when i assume k=n is true and i want to prove k=n+1 from k=n
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Now you assume true up to some \(n\in\mathbb{Z}^+\). So \[(x^n1)=(x1)(x^{n1}+x^{n2}+...+x+1)\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Now, look at \[(x1)(x^n+x^{n1}+x^{n2}+...+x+1).\]This is equal to \[(x1)x^n+(x1)(x^{n1}+x^{n2}+...+x+1)\]which by our inductive hypothesis, is equal to \[(x1)x^n+(x^n1)=x^{n+1}x^n+x^n1=x^{n+1}1.\]
 one year ago

johnny0929Best ResponseYou've already chosen the best response.0
ok that makes a lot of sense. thank you so much!
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
oh i read the exponent wrong, sorry
 one year ago
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