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Math Help - Proof

  1. #1
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    Proof

    Prove by induction that

    2( 4^{2n+1}) + 3^{3n+1} <br />
    is divisible by 11  \forall n \! \in \! \mathbb{N}
    ..
    Last edited by bobak; February 26th 2008 at 08:27 AM.
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  2. #2
    MHF Contributor kalagota's Avatar
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    are you sure?
    for n = 1, 4^{2(1)+1} + 3^{3(1)+1} = 4^3 + 3^4 = 145
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  3. #3
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    Sorry the latex was a bit confusing, I cleaned it up.
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  4. #4
    Super Member PaulRS's Avatar
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    Ok, let's assume it holds for n, we'll now show it does so for n+1

    a_{n}=2\cdot{4^{2n+1}}+3^{3n+1} is divisible by 11

    a_{n+1}=2\cdot{4^{2n+3}}+3^{3n+4}

    a_{n}-2\cdot{4^{2n+1}}=3^{3n+1} then 3^{3}\cdot(a_{n}-2\cdot{4^{2n+1}})=3^{3n+4}

    So: a_{n+1}=2\cdot{4^{2n+3}}+27\cdot{a_{n}}-54\cdot{4^{2n+1}}

    Since 2\cdot{4^{2n+3}}-54\cdot{4^{2n+1}}=2\cdot{4^{2n+1}}\cdot{(16-27)}=-2\cdot{4^{2n+1}}\cdot{11} (divisible by 11)

    and a_{n}=2\cdot{4^{2n+1}}+3^{3n+1} is divisible by 11

    It follows that a_{n+1}=2\cdot{4^{2n+3}}+3^{3n+4} is also divisible by 11
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