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Thread: Stuck on proof by induction

  1. #1
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    Stuck on proof by induction

    Hey, Stuck on this proof by induction question. I can get so far but not sure where to go after.

    $\displaystyle 3*2^n < n!$ when $\displaystyle n>4$

    Basis for Induction

    $\displaystyle 3*2^5 < 5!$

    $\displaystyle 96 < 120$

    Induction Step

    We assume $\displaystyle 3*2^n <n!$ when $\displaystyle n>4$

    We need to prove that $\displaystyle 3*2^n^+^1 < (n+1)!$

    $\displaystyle (n+1)! = (n+1)n!$

    $\displaystyle (3*2^n)(n+1)<(n+1)n!$

    $\displaystyle (3*2^n)(n+1)<(n+1)!$

    This is as far as I can get Any help would be appreciated!
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  2. #2
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    Quote Originally Posted by dtwazere View Post
    Hey, Stuck on this proof by induction question. I can get so far but not sure where to go after.

    $\displaystyle 3*2^n < n!$ when $\displaystyle n>4$

    Basis for Induction

    $\displaystyle 3*2^5 < 5!$

    $\displaystyle 96 < 120$

    Induction Step

    We assume $\displaystyle 3*2^n <n!$ when $\displaystyle n>4$

    We need to prove that $\displaystyle 3*2^n^+^1 < (n+1)!$

    $\displaystyle (n+1)! = (n+1)n!$

    $\displaystyle (3*2^n)(n+1)<(n+1)n!$

    $\displaystyle (3*2^n)(n+1)<(n+1)!$

    This is as far as I can get Any help would be appreciated!

    $\displaystyle 3\cdot 2^{n+1}=2(3\cdot 2^n)<2\cdot n! < (n+1)n!=(n+1)!$ , the last inequality being true since $\displaystyle n+1>2$ ...

    Tonio
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  3. #3
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    Recall that $\displaystyle n>1$ then $\displaystyle n+1>2$.

    Start with $\displaystyle 3\cdot 2^{n+1}=[3\cdot 2^{n}](2)<[n!](?)$
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