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Math Help - Proof by induction! Urgh! :(

  1. #1
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    Proof by induction! Urgh! :(

    edit
    11^2 + 12^1 = 133
    11n+3 + 12^2n+3
    But then I get stuck!? Any ideas?
    Last edited by Jimbobobo; November 21st 2007 at 01:02 AM.
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  2. #2
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    Say it is true for K>1, \left( {11} \right)^{K + 2}  + \left( {12} \right)^{2K + 1}  = 133j.
    Look at the case of K+1.
    \left( {11} \right)^{\left( {K + 1} \right) + 2}  + \left( {12} \right)^{2\left( {K + 1} \right) + 1}
      = \left( {11} \right)^{K + 3}  + 11\left( {12} \right)^{2K + 1}  - 11\left( {12} \right)^{2K + 1}  + \left( {12} \right)^{2K + 3}
    \begin{array}{l}<br />
  = 11\left[ {\left( {11} \right)^{K + 2}  + \left( {12} \right)^{2K + 1} } \right] + \left( {12} \right)^{2K + 1} \left[ { - 11 + \left( {12} \right)^2 } \right] \\ <br />
  = 11\left[ {133j} \right] + \left( {12} \right)^{2K + 1} \left[ {133} \right] \\ <br />
 \end{array}<br />
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  3. #3
    Forum Admin topsquark's Avatar
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    Plato either wins the award for most psychic Math Helper, or Jimbobobo edited his original question to the point where it is meaningless. I'm curious as to which?

    -Dan
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  4. #4
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    The original question was edited.
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