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Math Help - Proove by Induction

  1. #1
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    Proove by Induction

    I need to Prove that 6^(2n) -1 is divisible by 35. Using mathematical induction.

    Hint given : use the equality 6^2n = 36^n = ( 35+1)^n

    So step one is to verify if this works for 1, and so I did 6^(2*1) - 1 = 36-1=35 so it works.

    In step two I need to assume that this is true for an arbitrary n,

    Finally in the deductive proof I need to show that this holds for n+1.

    So I set it up as follows (35+1)^(n+1) - 1 = 35k, where k is some constant.
    I don't know where to go from here.
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  2. #2
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    Quote Originally Posted by meg0529 View Post
    I need to Prove that 6^(2n) -1 is divisible by 35. Using mathematical induction.

    Hint given : use the equality 6^2n = 36^n = ( 35+1)^n

    So step one is to verify if this works for 1, and so I did 6^(2*1) - 1 = 36-1=35 so it works.

    In step two I need to assume that this is true for an arbitrary n,

    Finally in the deductive proof I need to show that this holds for n+1.

    So I set it up as follows (35+1)^(n+1) - 1 = 35k, where k is some constant.
    I don't know where to go from here.
    (35 + 1)^{n + 1} - 1 = (35 + 1)(35 + 1)^n  - 1

     = 35(35 + 1)^n + (35 + 1)^n - 1


    Keeping in mind that you've assumed (35 + 1)^n - 1 is divisible by 35. So we can say it equals 35k, where k is some other whole number.

    Therefore,

    (35 + 1)^{n + 1} - 1 = 35(35 + 1)^n + 35k

     = 35[(35 + 1)^n + k]

    which is divisible by 35.

    QED.
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  3. #3
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    Ok the second half makes sense, I don't follow how you got this

     = 35(35 + 1)^n + (35 + 1)^n - 1

    Thanks!
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  4. #4
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    Quote Originally Posted by meg0529 View Post
    Ok the second half makes sense, I don't follow how you got this

     = 35(35 + 1)^n + (35 + 1)^n - 1

    Thanks!
    Distributive law.

    (35 + 1)(35 + 1)^n - 1 = 35(35 + 1)^n + 1(35 + 1)^n - 1.
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  5. #5
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    O doh! of course thanks so much.
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