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Thread: Induction help

  1. #1
    Junior Member
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    Exclamation Induction help

    I am stuck on this proof, but this is what i have done so far.


    When n=1 it is ture as (1+a)^1>= 1+a

    we can know assume P(k) is true:

    (1+a)^k>= 1+Ka

    so for P(K+1)= (1+a)^k+1>= (1+Ka)(1+K)
    >= (1+k+ka+k^2a)


    this looks wrong to me and i cannot figure it out on how to finish this proof can so one please help me.

    Thanks
    Last edited by nerdo; Jan 14th 2009 at 02:09 PM.
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  2. #2
    MHF Contributor

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    Start with $\displaystyle \left( {1 + a} \right)^K \geqslant 1 + Ka$ being true.
    Then go to the next step:
    $\displaystyle \begin{array}{rcl}
    {\left( {1 + a} \right)^{K + 1} } & = & {\left( {1 + a} \right)^K \left( {1 + a} \right)} \\
    {} & \geqslant & {\left( {1 + Ka} \right)\left( {1 + a} \right)} \\
    {} & = & {1 + a + Ka + Ka^2 } \\
    {} & > & {1 + a + Ka,\,\,({\color{blue}Ka^2 > 0)}} \\
    {} & = & {1 + \left( {K + 1} \right)a} \\ \end{array} $
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