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Math Help - sequences

  1. #1
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    sequences

    Find the sum to n terms of the sequence 100-2^2 +108-4^2 +116-6^2+ ...
    All help is much appreciated. Thank you
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  2. #2
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    Quote Originally Posted by Webby View Post
    Find the sum to n terms of the sequence 100-2^2 +108-4^2 +116-6^2+ ...
    This appears to be the subtraction of two different series:

    . . . . . 100\, +\, 108\, +\, 116\, +\, ...\, =\, \sum_{i=0}^{n-1}\, 100\, +\, 8i

    . . . . . 2^2\, +\, 4^2\, +\, 6^2\, +\, ...\, =\, \sum_{i=0}^{n-1}\, \left(2(i\, +\, 2)\right)^2

    What formulas do you have for these sorts of series?

    When you reply, please include a clear listing of your thoughts and working so far. Thank you!
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  3. #3
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    Quote Originally Posted by stapel View Post
    This appears to be the subtraction of two different series:

    . . . . . 100\, +\, 108\, +\, 116\, +\, ...\, =\, \sum_{i=0}^{n-1}\, 100\, +\, 8i

    . . . . . 2^2\, +\, 4^2\, +\, 6^2\, +\, ...\, =\, \sum_{i=0}^{n-1}\, \left(2(i\, +\, 2)\right)^2

    What formulas do you have for these sorts of series?

    When you reply, please include a clear listing of your thoughts and working so far. Thank you!
    The second should be:

    2^2\, +\, 4^2\, +\, 6^2\, +\, ...\, =\, \sum_{i=0}^{n-1}\, \left(2(i+1)\right)^2

    CB
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  4. #4
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    Sorry for taking so long, i have had electricity issues .
    To find the sum to n terms for a sequence such as this i have been using sigma notation.
    My current workings;
    100 + 108 + 116 +... to n/2 terms, -(2^2+4^2+6^2+...) to n/2 terms
    (8n+92) -(2n)^2

    8(n/2 sigma r=1)r + (n/2 sigma r=1)92 - 4(n/2 sigma r=1)r^2

    n(n+2) + 46n - n(n+2)(n+1)/6

    Substituting a values for n into this equation is not giving me the sum to n terms. Hence i have done something wrong up to this point.
    If you cannot understand anything please ask.
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  5. #5
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    Quote Originally Posted by Webby View Post
    Sorry for taking so long, i have had electricity issues .
    To find the sum to n terms for a sequence such as this i have been using sigma notation.
    My current workings;
    100 + 108 + 116 +... to n/2 terms, -(2^2+4^2+6^2+...) to n/2 terms
    (8n+92) -(2n)^2

    8(n/2 sigma r=1)r + (n/2 sigma r=1)92 - 4(n/2 sigma r=1)r^2

    n(n+2) + 46n - n(n+2)(n+1)/6

    Substituting a values for n into this equation is not giving me the sum to n terms. Hence i have done something wrong up to this point.
    If you cannot understand anything please ask.
    Could you restructure this more systematically so that we can understand what you have tried to do.

    (also, this would only work for n even, you will have to treat n odd seperatley)

    CB
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  6. #6
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    Quote Originally Posted by CaptainBlack View Post
    Could you restructure this more systematically so that we can understand what you have tried to do.

    (also, this would only work for n even, you will have to treat n odd seperatley)

    CB
    If instead of having it to n/2 terms, i had it to n terms. Would that make the sum to n terms work for all real numbers? This could be my mistake.

    My current workings;
    100 + 108 + 116 +... to n/2 terms, -(2^2+4^2+6^2+...) to n/2 terms
    (8n+92) -(2n)^2

    8(n/2 sigma r=1)r + (n/2 sigma r=1)92 - 4(n/2 sigma r=1)r^2

    using r = n(n+1)/2 r^2 = n(n+1)(2n+1)/6

    8 * (n/2)((n/2)+1)/2 + 92n/2 + 4 * (n/2)((n/2)+1)((2n/2)+1)/6

    n(n+2) + 46n - n(n+2)(n+1)/6
    Last edited by Webby; April 2nd 2009 at 01:47 AM.
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