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Math Help - Sum of the series

  1. #1
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    Sum of the series

    Please help me to solve this problem

    Find the sum of the series up to 2n terms
    1^2-2^2+3^2-4^2+5^2-7^2+.....
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  2. #2
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    Re: Sum of the series

    Quote Originally Posted by blackhat123 View Post
    Please help me to solve this problem

    Find the sum of the series up to 2n terms
    1^2-2^2+3^2-4^2+5^2-7^2+.....

    Are you missing a 6^2 term?
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  3. #3
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    Re: Sum of the series

    Ya I missed it out.

    Thanks for pointing it.
    The problem should be like this
    Find the sum of the series up to 2n terms
    1^2-2^2+3^2-4^2+5^2-6^2+.....
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  4. #4
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    Re: Sum of the series

    Here's one way it could be solved:

    \sum_{x=1}^{n}(2 x-1)^2 = \frac{1}{3} (4 n^3-n)
    \sum_{x=1}^{n}(2 x)^2= \frac{2}{3} n (n+1) (2 n+1)
    \sum_{x=1}^{n}(2 x-1)^2-\sum_{x=1}^{n}(2 x)^2 =\frac{1}{3} (4 n^3-n)-\frac{2}{3} n (n+1) (2 n+1)= -2 n^2-n

    That gives you 2 terms for each n, or 2n terms. I'm not certain this is how it's *supposed* to be solved, though
    Last edited by mr fantastic; June 14th 2011 at 01:30 AM.
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  5. #5
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    Re: Sum of the series

    Thanks Stro.So this is the sum of the series upto the first n terms.Now should I replace n by 2n to get the sum of the series upto 2n terms?Please guide me.
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  6. #6
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    Re: Sum of the series

    Edit: To answer your question: No, the above is the summation up to 2n terms within the original series, it would not be changed at this point, but you would have to show your work I bet.

    ie. \sum_{x=1}^{n}(2x)^2=4\cdot\sum_{x=1}^{n}x^2= 4\cdot\frac{n(n+1)(2n+1)}{6}=\frac{2}{3}n(n+1)(2n+  1)
    Spoiler:
    \sum_{x=1}^{n}(2 x-1)^2
    \sum_{x=1}^{n}4x^2-4x+1
    4\cdot\frac{n(n+1)(2n+1)}{6}-4\cdot\frac{n(n+1)}{2}+n
    \frac{4n^3+6n^2+2n}{3}-\frac{6n^2+6n}{3}+n
    \frac{1}{3}(4n^3+6n^2+2n-6n^2-6n+3n)
    \frac{1}{3} (4 n^3-n)
    Last edited by Stro; June 14th 2011 at 03:34 AM. Reason: Deleted extraneous info, see ivy's post
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  7. #7
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    Re: Sum of the series

    The sum of squares, for 1 to n, is well known to be (1/6)n(n+1)(2n+1) so the sum of squares from 1 to 2n is (1/6)(2n)(2n+1)(4n+1)= (1/3)n(2n+1)(4n+1).

    Here, however, we have a negative sign on all even squares so we have to subtract off 4+ 16+ ...+ (2n)^2 twice.
    But 4+ 16+ ...+ (2n)^2= 4(1)+ 4(4)+ ... 4(n^2)= 4(1+ 4+ ...+ n^2). Using the formula for sum of squares as before, that is 4(1/6)n(n+1)(2n+1)= (2/3)n(n+1)(2n+1). Notice that this is 1 to n, not 2n, now. Since we have to subtract that off twice, the original sum is
    (1/3)n(2n+1)(4n+1)- (4/3)n(n+1)(2n+1)= (1/3)n(2n+1)(4n+1- 4n-4)= (1/3)n(2n+1)(-3)= -n(2n+1).

    Check: If n= 3, 1- 4+ 9- 16+ 25- 36= -21
    -3(2(3)+1)= -3(7)= -21.
    Last edited by HallsofIvy; June 14th 2011 at 03:40 AM.
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  8. #8
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    Re: Sum of the series

    Thanks all for helping me out.
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