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Math Help - sum of multiples from 3 to 99

  1. #1
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    sum of multiples from 3 to 99

    If k=1 is the lower limit and n represents the upper limit and in the middle of the sigma we have (a+(k-1)d) to get the sum of multiples from 3 to 99 confuses me a quite a bit. I know we start at 3 but I am just guessing after that.
    Thank You,
    Keith stevens
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  2. #2
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    Quote Originally Posted by kcsteven View Post
    If k=1 is the lower limit and n represents the upper limit and in the middle of the sigma we have (a+(k-1)d) to get the sum of multiples from 3 to 99 confuses me a quite a bit. I know we start at 3 but I am just guessing after that.
    Thank You,
    Keith stevens
    You want the sum of the series 3, 6, 9, ..., 96, 99 I presume?

    The kth term of the series is 3 + 3(k - 1) where k runs from 1 to 33. So this is:
    \sum_{k = 1}^{33}[ 3 + 3(k - 1)] = 3 \sum_{k = 1}^{33}[1 + k - 1]

     = 3 \sum_{k = 1}^{33}k

    Now, \sum_{k = 1}^n k = \frac{n(n+1)}{2} so your sum is:
    \sum_{k = 1}^{33}[ 3 + 3(k - 1)] = 3 \cdot \frac{33(33 + 1)}{2} = 1683

    -Dan
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