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Math Help - polya strikes out

  1. #1
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    polya strikes out

    Write down the numbers 1,2,3,4,... in a row Srike out every third
    so
    we are left with 1,2,4,5,7,8,10... now write their cumulative sums
    of what remains. We are given
    1,3,7,12,19,27,37... Now strike out every second
    term then add the sums. So we are left with 8,27,64,125,216,...
    The pattern is 2^3,3^3,4^3,5^3,...

    How can I explain how this happens? I believe I have to write a formula for each step that I do. I do not know if this is right or how I could get the formula. Any help would be very nice. Thank you for your time.
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  2. #2
    Grand Panjandrum
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    If your previous post has not been answered it is either because no
    one knows how to answer it. Posting it again does not help, perhaps
    if you posted some of your thoughts in the original thread you would
    get more help.

    RonL
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  3. #3
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    Quote Originally Posted by schinb64 View Post
    Write down the numbers 1,2,3,4,... in a row Srike out every third
    so
    we are left with 1,2,4,5,7,8,10... now write their cumulative sums
    of what remains. We are given
    1,3,7,12,19,27,37... Now strike out every second
    term then add the sums. So we are left with 8,27,64,125,216,...
    The pattern is 2^3,3^3,4^3,5^3,...

    How can I explain how this happens? I believe I have to write a formula for each step that I do. I do not know if this is right or how I could get the formula. Any help would be very nice. Thank you for your time.
    It seems like a long problem.

    But it does help if you have a formula.

    Let {b_n} be the sequence the represents after striking out every third one. Then, b_n=n+[(n-1)/2].
    Where [ ] means "greatest integer function".

    Okay, the next step is to find the sum of all terms before.

    Thus,

    SUM(k=1,n) b_k = SUM(k=1,n) k+[(k-1)/2]

    To find this use the fact that,

    SUM(k=1,n) k = n(n+1)/2

    And

    SUM(k=1,n) [k/2] = n^2 if n is even, otherwise n^2+n.
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