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johnny0929

  • 3 years ago

prove by induction that ((x^(k))-1)=(x-1)(x^(k-1)+x^(k-2)+...+x+1) i got stuck when i have to prove k=n+1 from k=n being assumed true

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  1. anonymous
    • 3 years ago
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    something is wrong with that formula

  2. johnny0929
    • 3 years ago
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    what is wrong with the formula?

  3. KingGeorge
    • 3 years ago
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    Looks correct to me.

  4. KingGeorge
    • 3 years ago
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    Anyways, you have to start with a base case of k=1. Then, you have\[(x-1)\overset{\text{?}}{=} (x-1)(1)\]Which is true, so the base case is good.

  5. johnny0929
    • 3 years ago
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    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

  6. KingGeorge
    • 3 years ago
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    Now you assume true up to some \(n\in\mathbb{Z}^+\). So \[(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+...+x+1)\]

  7. KingGeorge
    • 3 years ago
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    Now, look at \[(x-1)(x^n+x^{n-1}+x^{n-2}+...+x+1).\]This is equal to \[(x-1)x^n+(x-1)(x^{n-1}+x^{n-2}+...+x+1)\]which by our inductive hypothesis, is equal to \[(x-1)x^n+(x^n-1)=x^{n+1}-x^n+x^n-1=x^{n+1}-1.\]

  8. KingGeorge
    • 3 years ago
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    That make sense?

  9. johnny0929
    • 3 years ago
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    ok that makes a lot of sense. thank you so much!

  10. KingGeorge
    • 3 years ago
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    You're welcome.

  11. anonymous
    • 3 years ago
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    oh i read the exponent wrong, sorry

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