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Thread: Proof

  1. #1
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    Proof

    Prove by induction that

    $\displaystyle 2( 4^{2n+1}) + 3^{3n+1}
    $
    is divisible by 11 $\displaystyle \forall n \! \in \! \mathbb{N}$
    ..
    Last edited by bobak; Feb 26th 2008 at 07:27 AM.
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  2. #2
    MHF Contributor kalagota's Avatar
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    are you sure?
    for n = 1, $\displaystyle 4^{2(1)+1} + 3^{3(1)+1} = 4^3 + 3^4 = 145$
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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

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

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

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

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

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

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

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