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Math Help - Prove By Mathematical Induction

  1. #1
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    Prove By Mathematical Induction

    I'm stuck on this question:

    State the principle of mathematical induction, and use it to prove that for n \geq 4

    4n! > 2^{n+2}

    So far I've done this:

    P(n) is a statement for each natural number n, where n \geq 4. P(n) is true if:

    • P(n) is true for n = 4.
    • If it is true for arbitrary n, it is also true for n + 1.


    For n = 4, 4(4!) > 2^{6}, 96 > 64 hence true.

    For n + 1:

    4(n + 1)! > 2^{n + 3}

    (n + 1)! > 2^{n + 1}

    n!(n+1) > 2^{n}(n+1) (Using hypothesis)

    And that's where I get stuck. RHS should be 2^{n+1}, but that only occurs when n = 1.

    Please help!
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  2. #2
    MHF Contributor
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    Quote Originally Posted by splbooth View Post
    I'm stuck on this question:

    State the principle of mathematical induction, and use it to prove that for n \geq 4

    4n! > 2^{n+2}

    So far I've done this:

    P(n) is a statement for each natural number n, where n \geq 4. P(n) is true if:

    • P(n) is true for n = 4.
    • If it is true for arbitrary n, it is also true for n + 1.


    For n = 4, 4(4!) > 2^{6}, 96 > 64 hence true.

    For n + 1:

    4(n + 1)! > 2^{n + 3}

    (n + 1)! > 2^{n + 1}

    n!(n+1) > 2^{n}(n+1) (Using hypothesis)

    And that's where I get stuck. RHS should be 2^{n+1}, but that only occurs when n = 1.

    Please help!
    you can try to prove whether or not 4(n+1)!>2^{(n+1)+2} if 4n!>2^{n+2}

    P(k)

    4k!>2^{k+2}

    P(k+1)

    4(k+1)!>2^{(k+1)+2}

    Proof

    4(k+1)!=4(k+1)k!=(k+1)4k!

    If k\ge4 then (k+1)4k!\ge(5)4k!

    2^{k+3}=(2)2^{k+2}

    Then if 4k!>2^{k+2}

    4(k+1)!>2^{k+3} certainly since (5)4k!>(2)2^{k+2}
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  3. #3
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    Smile

    That's great, thanks
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