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Math Help - Another proof question

  1. #1
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    Another proof question

    I want to prove the following theorem:
    ----------------------------------------
    If n is a positive integer, then 7^n-1 is divisible by 6.
    ----------------------------------------
    The only idea I have so far is the following equation:

    \frac{1}{6}(7^n-1)=a_n

    Where a_n is an infinite sequence:

    a_n={1,8,57,400,2801,...}

    The sequence doesn't have a common ratio, however I have observed the following:

    8:1=8
    57:8=7.125
    400:57\approx 7.02
    2801:400=7.0025

    So it seems that \frac{1}{6}\lim_{n->\infty}\frac{7^n-1}{7^{n-1}-1}=7

    I'm not really sure if the limit is useful or not.
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  2. #2
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    Hello, adkinsjr!

    Prove the following:

    If n is a positive integer, then 7^n-1 is divisible by 6.
    Consider the expansion of (6+1)^n

    . . 7^n \;\;=\;\;(6\;+\;1)^n \;=\; \;6^n \;+\; {n\choose1}6^{n-1} \;+\; {n\choose2}6^{n-2} \;+ \hdots +\; {n\choose n-2}6^2 \;+\; {n\choose n-1}6 \;+\; 1


    Then: . 7^n - 1 \;\;=\;\;6^n + {n\choose1}6^{n-1} + {n\choose2}6^{n-2} + \hdots + {n\choose n-2}6^2 + {n\choose n-1}6

    . . . . . . . . . . =\;\;6\cdot\underbrace{\bigg[6^{n-1} + {n\choose1}6^{n-2} + {n\choose2}6^{n-3} + \hdots + {n\choose n-2}6 + {n\choose n-1}\bigg]}_{\text{This is an integer}}

    Hence, 7n - 1 is a multiple of 6.


    Therefore, 7^n-1 is divisible by 6.

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, adkinsjr!

    Consider the expansion of (6+1)^n

    . . 7^n \;\;=\;\;(6\;+\;1)^n \;=\; \;6^n \;+\; {n\choose1}6^{n-1} \;+\; {n\choose2}6^{n-2} \;+ \hdots +\; {n\choose n-2}6^2 \;+\; {n\choose n-1}6 \;+\; 1


    Then: . 7^n - 1 \;\;=\;\;6^n + {n\choose1}6^{n-1} + {n\choose2}6^{n-2} + \hdots + {n\choose n-2}6^2 + {n\choose n-1}6

    . . . . . . . . . . =\;\;6\cdot\underbrace{\bigg[6^{n-1} + {n\choose1}6^{n-2} + {n\choose2}6^{n-3} + \hdots + {n\choose n-2}6 + {n\choose n-1}\bigg]}_{\text{This is an integer}}

    Hence, 7n - 1 is a multiple of 6.


    Therefore, 7^n-1 is divisible by 6.
    Ok, I understand this. But how did you develop the idea of expanding the binomial? When ever I go to prove theorems like this, I'm not sure where to begin. You're proof just isn't something I would have noticed immediately.

    What particular branch of mathematics should I study in order to obtain this skill?
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  4. #4
    Newbie Pi R Squared's Avatar
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    Post Another Proof Question

    I am not the one that helped you with your proof, but I am a Mathematics major. Depending on how far you plan to go with developing proofs, you may want to take Discrete Math. If you plan to pursue a career in Mathematics you may want to take something like Foundations of Mathematics. The college you attend may call it something else, but it will be a class strictly on writing proofs. Talk to your adviser and find out what the best classes to take relating to your major.


    Hope you find what you are looking for!
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by adkinsjr View Post
    I want to prove the following theorem:
    ----------------------------------------
    If n is a positive integer, then 7^n-1 is divisible by 6.
    ----------------------------------------
    The only idea I have so far is the following equation:

    \frac{1}{6}(7^n-1)=a_n

    Where a_n is an infinite sequence:

    a_n={1,8,57,400,2801,...}

    The sequence doesn't have a common ratio, however I have observed the following:

    8:1=8
    57:8=7.125
    400:57\approx 7.02
    2801:400=7.0025

    So it seems that \frac{1}{6}\lim_{n->\infty}\frac{7^n-1}{7^{n-1}-1}=7

    I'm not really sure if the limit is useful or not.
    It's useful to know that a^n-b^n=(a-b)\cdot(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+ab^{n-2}+b^{n-1}), since if you multiply out it's easy to see that all terms cancel except a\cdot a^{n-1} and -b\cdot b^{n-1}.
    This can be applied to your problem directly like this 7^n-1=7^n-1^n=(7-1)\cdot(7^{n-1}+7^{n-2}\cdot 1+\cdots+7\cdot 1^{n-2}+1^{n-1})
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