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Thread: 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 $\displaystyle n \geq 4$

    $\displaystyle 4n! > 2^{n+2}$

    So far I've done this:

    $\displaystyle P(n)$ is a statement for each natural number $\displaystyle n$, where $\displaystyle n \geq 4$. $\displaystyle P(n)$ 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 $\displaystyle n = 4$, $\displaystyle 4(4!) > 2^{6}$, $\displaystyle 96 > 64$ hence true.

    For $\displaystyle n + 1$:

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

    $\displaystyle (n + 1)! > 2^{n + 1}$

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

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

    Please help!
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  2. #2
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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 $\displaystyle n \geq 4$

    $\displaystyle 4n! > 2^{n+2}$

    So far I've done this:

    $\displaystyle P(n)$ is a statement for each natural number $\displaystyle n$, where $\displaystyle n \geq 4$. $\displaystyle P(n)$ 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 $\displaystyle n = 4$, $\displaystyle 4(4!) > 2^{6}$, $\displaystyle 96 > 64$ hence true.

    For $\displaystyle n + 1$:

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

    $\displaystyle (n + 1)! > 2^{n + 1}$

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

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

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

    P(k)

    $\displaystyle 4k!>2^{k+2}$

    P(k+1)

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

    Proof

    $\displaystyle 4(k+1)!=4(k+1)k!=(k+1)4k!$

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

    $\displaystyle 2^{k+3}=(2)2^{k+2}$

    Then if $\displaystyle 4k!>2^{k+2}$

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

    That's great, thanks
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