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Thread: Use induction

  1. #1
    MHF Contributor alexmahone's Avatar
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    Use induction

    Use induction to prove that $\displaystyle 3^n>n^4$ if $\displaystyle n\geq8$.

    My attempt:

    For n=8, the statement becomes 6561>4096, which is true.

    Assume that the statement is true for n.

    Then, $\displaystyle 3^{n+1}>3n^4$

    We need to prove that $\displaystyle 3n^4>(n+1)^4$

    $\displaystyle \Leftrightarrow 3n^4>n^4+4n^3+6n^2+4n+1$

    $\displaystyle \Leftrightarrow -2n^4+4n^3+6n^2+4n+1>0$ for $\displaystyle n\geq8$.

    What's the easiest way of proving this?

    Edit: The above inequality is obviously not true for large n. What should I do, instead?
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  2. #2
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    Re: Use induction

    Quote Originally Posted by alexmahone View Post
    Use induction to prove that $\displaystyle 3^n>n^4$ if $\displaystyle n\geq8$.

    My attempt:

    For n=8, the statement becomes 6561>4096, which is true.

    Assume that the statement is true for n.

    Then, $\displaystyle 3^{n+1}>3n^4$

    We need to prove that $\displaystyle 3n^4>(n+1)^4$

    $\displaystyle \Leftrightarrow 3n^4>n^4+4n^3+6n^2+4n+1$

    $\displaystyle \Leftrightarrow -2n^4+4n^3+6n^2+4n+1>0$ for $\displaystyle n\geq8$.

    What's the easiest way of proving this?

    Edit: The above inequality is obviously not true for large n. What should I do, instead?
    first of all, your last inequality is wrong. why?

    also, you shouldn't have expanded $\displaystyle (n+1)^4.$ just take the fourth root to get the inequality $\displaystyle \sqrt[4]{3} > 1 + \frac{1}{n},$ which is true for all $\displaystyle n \geq 4$ because $\displaystyle 1 + \frac{1}{n} \leq 1+ \frac{1}{4} = 1.25$ but $\displaystyle \sqrt[4]{3} > 1.3$.
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: Use induction

    Quote Originally Posted by NonCommAlg View Post
    first of all, your last inequality is wrong. why?
    Oops, it should have been $\displaystyle -2n^4+4n^3+6n^2+4n+1<0$.
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