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Last edited by SarahGr; May 14th 2009 at 08:26 PM.
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Originally Posted by SarahGr 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" ?
Originally Posted by SarahGr 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
Last edited by SarahGr; May 14th 2009 at 08:29 PM.
Originally Posted by SarahGr 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 where n is the number of terms, a is the first term and d is the common difference
Last edited by SarahGr; May 14th 2009 at 08:27 PM.
Originally Posted by SarahGr 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 , there is an easy way: group the terms 2 by 2, you get . As you can see, all the parentheses equal -2, and there are ?? of them, so...
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