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Math Help - 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; January 14th 2009 at 03:09 PM.
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  2. #2
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    Start with \left( {1 + a} \right)^K  \geqslant 1 + Ka being true.
    Then go to the next step:
    \begin{array}{rcl}<br />
   {\left( {1 + a} \right)^{K + 1} } &  =  & {\left( {1 + a} \right)^K \left( {1 + a} \right)}  \\<br />
   {} &  \geqslant  & {\left( {1 + Ka} \right)\left( {1 + a} \right)}  \\<br />
   {} &  =  & {1 + a + Ka + Ka^2 }  \\<br />
   {} &  >  & {1 + a + Ka,\,\,({\color{blue}Ka^2  > 0)}}  \\<br />
   {} &  =  & {1 + \left( {K + 1} \right)a}  \\ \end{array}
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