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Math Help - Finding the sum of a series

  1. #1
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    Finding the sum of a series

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    Last edited by SarahGr; May 14th 2009 at 07:26 PM.
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    Quote Originally Posted by SarahGr View Post
    Find the sum of the series 1-3+5-7+-11+...+1001?

    I have no idea on this one....

    I do know that the sum formula is Sn=n(t1+tn)/2
    1 - 3 + 5 - 7 + ... - 995 + 997 - 999 + 1001 = ?

    sum, Gauss style ...

    (1001 + 1) - (999 + 3) + (997 + 5) - (995 + 7) + ...

    now, you have to determine if the individual terms all match up (yielding a sum of zero), or is there a term "left over" ?
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    Quote Originally Posted by SarahGr View Post
    Find the sum of the series 1-3+5-7+-11+...+1001?

    I have no idea on this one....

    I do know that the sum formula is Sn=n(t1+tn)/2
    Is this supposed to be the alternating sign sum of the odd integers up to 1001?

    If so write it as the difference of two arithmetic progressions:

    (1 + 5 + 9 + ... + 1001) - (3 + 7 + 11 + ... + 999)

    CB
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    Last edited by SarahGr; May 14th 2009 at 07:29 PM.
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    Quote Originally Posted by SarahGr View Post
    Is there a formula that i can use to do those out? I'm not sure I understand what you mean
    The sum of an arithmetic sequence is S_n = \frac{n}{2}(2a+(n-1)d)

    where n is the number of terms, a is the first term and d is the common difference
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    Last edited by SarahGr; May 14th 2009 at 07:27 PM.
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    Quote Originally Posted by SarahGr View Post
    ahh I'm so sorry guys the sequence is actually 1-3+5-7+7-11+...+1001

    (I forgot the 7 =/)
    This "+7" would be weird; I can't see the pattern.

    For S=1-3+5-7+\cdots+1001, there is an easy way: group the terms 2 by 2, you get S=(1-3)+(5-7)+(9-11)+\cdots+(997-999)+1001. As you can see, all the parentheses equal -2, and there are ?? of them, so...
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    Last edited by SarahGr; May 14th 2009 at 07:27 PM.
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