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Thread: Study trouble with specific problem example

  1. #1
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    Question Study trouble with specific problem example

    I have an exam coming up on Wednesday and am still trying to decipher the logical problem-solving steps needed to solve a question by induction like this (from the practice questions pool):

    For all natural numbers n, 3^n < (n+3)!

    I have deciphered some of the problem solving into a couple base cases:

    case 1: n = 0, so 3^0 < (0+3)!, 1 < 6, so the base case works but I don't know where to go from here.
    case 2: n = 0 works as shown in the last example, and then the equation can be rewritten to get another base case to use induction.
    The rewritten equation is 3 < ((n+3)!)^1/n, just taking the nth root of both sides. By using n = 1 as a new base case and getting 3 < 24, I have shown that this method also works. But how can I manipulate this equation to prove that it keeps working? How can I use ((n + 1) + 3)! to show that this will always be getting bigger and thus will remain bigger than 3?
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  2. #2
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    Re: Study trouble with specific problem example

    Quote Originally Posted by oli1230 View Post
    I have an exam coming up on Wednesday and am still trying to decipher the logical problem-solving steps needed to solve a question by induction like this (from the practice questions pool):

    For all natural numbers n, 3^n < (n+3)!

    I have deciphered some of the problem solving into a couple base cases:

    case 1: n = 0, so 3^0 < (0+3)!, 1 < 6, so the base case works but I don't know where to go from here.
    case 2: n = 0 works as shown in the last example, and then the equation can be rewritten to get another base case to use induction.
    The rewritten equation is 3 < ((n+3)!)^1/n, just taking the nth root of both sides. By using n = 1 as a new base case and getting 3 < 24, I have shown that this method also works. But how can I manipulate this equation to prove that it keeps working? How can I use ((n + 1) + 3)! to show that this will always be getting bigger and thus will remain bigger than 3?
    Well just look at THIS
    Once true it is always true.
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  3. #3
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    Re: Study trouble with specific problem example

    Is that really how it goes? I am not missing any specific identifiers or procedures?
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    MHF Contributor MarkFL's Avatar
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    Re: Study trouble with specific problem example

    If I were going to prove the given hypothesis using induction, I would begin with the base case $\displaystyle P_1$ (we begin with $\displaystyle n=1$ as 1 is generally taken to be the smallest natural number):

    $\displaystyle 3<(1+3)!=4!$

    $\displaystyle 3<24$

    This is true, and so the induction hypothesis $\displaystyle P_n$ is:

    $\displaystyle 3^n<(n+3)!$

    In order to determine the induction step, I would look at $\displaystyle 3P_n$:

    $\displaystyle 3\cdot3^n<3(n+3)!$

    Or:

    $\displaystyle 3^{n+1}<3(n+3)!$

    Now, for all $\displaystyle n\in\mathbb{N}$ we must have:

    $\displaystyle 0<n$

    Or:

    $\displaystyle 3<n+3$

    Hence:

    $\displaystyle 3<(n+1)+3$

    Thus:

    $\displaystyle 3(n+3)!<((n+1)+3)(n+3)!=((n+1)+3)!$

    And so we have:

    $\displaystyle 3^{n+1}<3(n+3)!<((n+1)+3)!$

    Or:

    $\displaystyle 3^{n+1}<((n+1)+3)!$

    We have derived $\displaystyle P_{n+1}$ from $\displaystyle P_n$, thereby completing the proof by induction.
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