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Math Help - Problem in series

  1. #1
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    Problem in series

    Find the sum of the series

    1^2 2^2 + 3^2 4^2 + - (2n)^2

    How can I solve this?
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  2. #2
    Member SengNee's Avatar
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    Pangkor Island, Perak, Malaysia.
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    You can find out the solution here.
    http://www.mathhelpforum.com/math-he...10-series.html
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  3. #3
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    As we know,
    nth term = a + (n - 1)d

    so: (4n-1) = 1 4n

    There Ive confusion.

    If LHS=RHS, whats they mean?

    We need to do summation of (1-4n) only? Or nth term is (1-4n)?

    But in the arithmetic series we know summation of the series is Sn = n/2 [2a + (n - 1)d]

    By following arithmetic formulae I didnt get the correct answer.

    But If
    n
    (1-4n) = -n(2n + 1)
    r=1

    And this is the correct answer.
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