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Math Help - [SOLVED] Proof divisible by 64

  1. #1
    Tatanka
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    [SOLVED] Proof divisible by 64

    Can someone point me in the right direction of solving the following problem:

    Prove that for any postive integer n, the value of the expression 3^(2n+2)-8n-9 is divisible by 64.
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  2. #2
    hpe
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    Quote Originally Posted by Tatanka
    Can someone point me in the right direction of solving the following problem:

    Prove that for any postive integer n, the value of the expression 3^(2n+2)-8n-9 is divisible by 64.
    Use mathematical induction.

    Set x_n = 3^(2n+2) -8n -9.
    Solve for the power term on the right:

    x_n + 8n + 9 = 3^(2n+2)

    Now look at the term for the next n:

    x_n+1 = 3^(2n+4) -8(n+1) - 9
    = 9*3^(2n+2) - 8n - 17
    = 9*(x_n + 8n + 9) -8n - 17
    = ...

    Use this to prove that if x_n is divisible by 64, x_n+1 is divisible by 64 as well.

    You still need to start the induction for some n:-)
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  3. #3
    vms
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    Quote Originally Posted by Tatanka
    Can someone point me in the right direction of solving the following problem:

    Prove that for any postive integer n, the value of the expression 3^(2n+2)-8n-9 is divisible by 64.


    3^(2n+2)-8n-9 =9^(n+1)-8n-9=(1+8)^(n+1)-8n-9

    expanding it by using binomial theorem,
    1+(n+1)8+C(n+1, 2)8^2+C(n+1, 3)8^3+C(n+1, 4)8^4....+C(n+1, n+1)8^n+1-8n-9

    first two terms get cancel with the last two terms and
    then each term contains 8^2 common ...
    hence the number remained is always divisible by 64
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