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Math Help - one step to the other..how?

  1. #1
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    one step to the other..how?

    im reviewing a proof by induction and wondering how

    2^{n+1} >2n^{2} = (n+1)^{2} + (n-2n-1)

    the original question was prove 2^{n} >n^{2} for all n  \geq 5

    thanks
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by skystar View Post

    2n^{2} = (n+1)^{2} + (n-2n-1)
    (n+1)^2=n^2+2n+1

    Hence (n+1)^{2} + (n-2n-1)=2n^2

    2^{n+1} >2n^{2}
    By multiplying the hypothesis : 2^n>n^2 by 2

    Does it help you understand ?
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  3. #3
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    but would that go to n^2 +n confused still
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  4. #4
    Moo
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    Quote Originally Posted by skystar View Post
    but would that go to n^2 +n confused still
    Where ? oO
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