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Math Help - Operation @

  1. #1
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    Operation @

    If operation @ is defned as: (1) 2 @ 2 = 1; (2) (2n+2) @ 2 = (2n @ 2) + (n)/2, what is 2010 @ 2?
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  2. #2
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    If you look at the series 2 @ 2, 4 @ 2, 6 @ 2, ...

    2 @ 2 = 1 = (1/2) * 2
    4 @ 2 = 2 @ 2 + 1/2 = 3/2 = (1/2) * (2 + 1)
    6 @ 2 = 4 @ 2 + 4/2 = 7/2 = (1/2) * (2 + 1 + 4)
    8 @ 2 = 6 @ 2 + 9/2 = 8 = (1/2) * (2 + 1 + 4 + 9)

    You might be interested in this formula: 1^2+2^2+3^2+.....+n^2=n*(n+1)*(2n+1)/6. I think you can get to the answer from there.

    Post again in this thread if you're still having trouble.
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  3. #3
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    I have been playing with this for several days and still cannot seem to put it together
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  4. #4
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    2 @ 2 = 1 = (1/2) * 2
    4 @ 2 = 2 @ 2 + 1/2 = 3/2 = (1/2) * (2 + 1)
    6 @ 2 = 4 @ 2 + 4/2 = 7/2 = (1/2) * (2 + 1 + 4) = (1/2) * (2 + (1 + 4))
    8 @ 2 = 6 @ 2 + 9/2 = 8 = (1/2) * (2 + 1 + 4 + 9) = (1/2) * (2 + (1 + 4 + 9))
    10 @ 2 = 8 @ 2 + 16/2 = (1/2) * (2 + 1 + 4 + 9 + 16) = (1/2) * (2 + (1 + 4 + 9 + 16))

    so the pattern is:

    2x @ 2 = (1/2) * (2 + (1 + 4 + 9 + ... + (x-1)^2))

    and you can use (with n=x-1):

    1^2+2^2+3^2+.....+n^2=n*(n+1)*(2n+1)/6

    and then set x = 1005 to get 2010 @ 2.

    - Hollywood
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  5. #5
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    Quote Originally Posted by hollywood View Post
    2 @ 2 = 1 = (1/2) * 2
    4 @ 2 = 2 @ 2 + 1/2 = 3/2 = (1/2) * (2 + 1)
    6 @ 2 = 4 @ 2 + 4/2 = 7/2 = (1/2) * (2 + 1 + 4) = (1/2) * (2 + (1 + 4))
    8 @ 2 = 6 @ 2 + 9/2 = 8 = (1/2) * (2 + 1 + 4 + 9) = (1/2) * (2 + (1 + 4 + 9))
    10 @ 2 = 8 @ 2 + 16/2 = (1/2) * (2 + 1 + 4 + 9 + 16) = (1/2) * (2 + (1 + 4 + 9 + 16))

    so the pattern is:

    2x @ 2 = (1/2) * (2 + (1 + 4 + 9 + ... + (x-1)^2))

    and you can use (with n=x-1):

    1^2+2^2+3^2+.....+n^2=n*(n+1)*(2n+1)/6

    and then set x = 1005 to get 2010 @ 2.

    - Hollywood
    Must you not check if the above formula is correct by induction??
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  6. #6
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    Yes, you should. I'm not sure whether you mean the sum-of-squares or the formula for 2x @ 2. But you can do both at once - just substitute the sum-of-squares formula into the other:

    2x @ 2 = (1/2) * (2 + (x-1)*(x)*(2x-1)/6)

    It should be pretty easy to do. Usually the difficult part is finding the formula - doing the proof is just working through the algebra.

    - Hollywood
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