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anonymous
 3 years 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
anonymous
 3 years 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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0something is wrong with that formula

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what is wrong with the formula?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Looks correct to me.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Anyways, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yup 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

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

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Now, 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 \[(x1)x^n+(x^n1)=x^{n+1}x^n+x^n1=x^{n+1}1.\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok that makes a lot of sense. thank you so much!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh i read the exponent wrong, sorry
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