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Math Help - mathematical induction

  1. #1
    baz
    baz is offline
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    mathematical induction


    Determine, with proof, for what natural numbers
    n the inequality

    n
    ! =<(n + 1)2^n

    is satisfied.

    It is true for 1=< n<=5, but Iam not able to prove it through procedure of mathematical induction.

    Will anybody give me hint.
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  2. #2
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    So it's false for n=6. If it's false for n, can you prove that it's false for n+1? In other words, given n! > (n+1)2^n, can you prove (n+1)! > (n+2)2^(n+1)?

    As a hint, since you have:
    n^2 > 3
    n^2+1 > 4
    n^2+2n+1 > 2n+4
    (n+1)^2 > 2(n+2)

    So, by induction, it's false for all n>5. And you have the result for 1 <= n <= 5, so that accounts for all the natural numbers.

    Post again in this thread if you're still having trouble.
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  3. #3
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    Quote Originally Posted by baz View Post

    Determine, with proof, for what natural numbers
    n the inequality

    n
    ! =<(n + 1)2^n

    is satisfied.

    It is true for 1=< n<=5, but Iam not able to prove it through procedure of mathematical induction.

    Will anybody give me hint.
    You can instead prove for n\ \ge\ 6

    n!\ \ge\ (n+1)2^n

    Try to prove that this causes

    (n+1)!\ \ge\ (n+2)2^{n+1}

    P(k)

    k!\ \ge\ (k+1)2^k

    P(k+1)

    (k+1)!\ \ge\ (k+2)2^{k+1}

    Obtain the P(k+1) statement in terms of P(k)

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

    2^{k+1}=(2)2^k

    (k+1)k!\ \ge\ 2(k+2)2^k ?

    Now all you need show is that if

    (k+1)k!\ \ge\ (k+1)(k+1)2^k

    then

    (k+1)k!\ \ge\ 2(k+2)2^k for k\ \ge\ 6

    so you need show

    (k+1)^2\ \ge 2(k+2) for k\ \ge\ 6

    This proves that your original inequality does not hold for any n above 5.
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  4. #4
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    Even though you are only dealing with a few values of n,
    you can complete the proof for your original inequality starting with n=5,
    by proving that the inequality must be true for the next n down,
    given that n is natural.

    P(k)

    k!\ \le\ (k+1)2^k ?

    P(k-1)

    (k-1)!\ \le\ k2^{k-1} ?

    Express this as much as possible using P(k).

    (k)(k-1)!\ \le\ (k)k2^{k-1}

    k!\ \le\ \frac{k^2}{2}2^k

    If k!\ \le\ (k+1)2^k then k!\ \le\ \frac{k^2}{2}2^k if \frac{k^2}{2}\ \ge\ (k+1)

    Hence

    k^2\ \ge\ 2k+2 ?

    k^2-2k-2\ \ge\ 0 ?

    The graph of this quadratic is U-shaped and the roots are complex,
    hence it is entirely above the x-axis

    \Rightarrow\ k^2-2k-2\ >0

    Therefore, if n!\ \le\ (n+1)2^n for some n=k, it is true for all n<k, for n natural.
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