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Find the sum of the series 1-3+5-7+-11+ +1001

  1. May 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the sum of the series 1-3+5-7+-11+...+1001


    2. Relevant equations

    I have no idea on this one....

    I do know that the sum formula is Sn=n(t1+tn)/2
     
  2. jcsd
  3. May 13, 2009 #2

    HallsofIvy

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    That formula only applies to arithmetic sequences and this is not an arithmetic sequence.

    The first thing you will need to do is clarify the sum: 1-3+5-7+-11+...+1001. the first numbers, in absolute value, 1, 3, 5, and 7 differ by 2 but then there is the jump to 11, 4 larger than 7. Is it supposed to be 1-3+5-7+9-11+...+1001? If so then you can rewrite it 1+ (5-3)+ (9-7)+ (13- 11)+ ...+ (1001-999)= 1+ 2+ 2+ 2+ ...+ 2. Now, how many "2"s are there? How many pairs of odd numbers are there from 5 to 1001?
     
  4. May 13, 2009 #3

    Yes sorry, there is supposed to be a 9 in there...

    So there would be 999 "2"s ?

    Would there be 200 pairs of odd numbers?
     
  5. May 20, 2009 #4
    an=a+(n-1)d
    1001=3+(n-1)2
    1001-3=(n-1)2
    998/2=n-1
    499=n-1
    499+1=n
    n=500

    500 odd integers from 3-1001

    Therefore,
    250 2s

    250*2=500
    500+1=501

    Therefore,
    the sum of the sequence=501 :wink:
     
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