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Math Help - maths - sum of a series

  1. #1
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    maths - sum of a series

    a. find the sum of the first 200 pos. intergers of
    1 + 2 + 3 + 4 + . . . + 200
    is it
    Tn =a+(n-1)d
    200=1+(n-1)1
    =1+n-1
    200=n
    n=200

    then
    Sn=n/2(a+l)
    S200=200/2(1+200)
    =100 X 201
    =20100
    is that right?

    b. series 1 + 5 + 7 + 11 + . . . + 199 is formed by omitting all those which are multiples of 2 or 3 up to the first 200 intergers. Find the sum

    how do you do this??
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  2. #2
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    Your first sum is correct. Think 100 pairs of numbers with each pair adding up to 201 (1+200, 2+199, 3+198, etc)

    The 2nd sum can be done in the same way. Add up all the numbers from 1 to 199. Subtract the series:
    2+4+6+8+...198
    Subtract the series
    3+6+9+12+...198
    Add back the series
    6+12+18+...198.
    I think this is called inclusion/exclusion.
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  3. #3
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    ok so you find the sum of the whole lot
    then minus the sum of the 2 multiples
    then minus the sum of the 3 multiples
    then add back the sum of the other
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