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Math Help - [SOLVED] induction with multiples

  1. #1
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    [SOLVED] induction with multiples

    how to prove this

    5^{2n}-3^n when n\geq0

    i do the base step of proving it works for 0

    5^{2(0)}-3^0

    =0 and 0\times11=0 zero is a multiple of 11

    so assume 5^{2n}-3^n now the inductive step using n+1. Where do i start?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Rapid_W View Post
    how to prove this

    5^{2n}-3^n is a multiple of 11 when n\geq0

    i do the base step of proving it works for 0

    5^{2(0)}-3^0

    =0 and 0\times11=0 zero is a multiple of 11

    so assume 5^{2n}-3^n now the inductive step using n+1. Where do i start?
    I added the part in bold...that seemed to be a very important part of the problem that was left out.

    Now, when you apply the inductive step, suppose for some k\in\mathbb{N} that 5^{2k}-3^k=11m for some m\in\mathbb{N}

    Now, it follows that

    5^{2(k+1)}-3^{k+1}=25\cdot5^{2k}-3\cdot 3^{k}=3\left(5^{2k}-3^{k}\right)+22\cdot5^{2k}=3\left(11m\right)+22\cd  ot5^{2k} =11\left(3m+2\cdot5^{k}\right), which is obviously a multiple of 11.

    Does this make sense?
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  3. #3
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    yes i can completly follow this thanks
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