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Math Help - This series

  1. #1
    Senior Member
    Joined
    Oct 2008
    Posts
    393

    This series

    10 + (10+3) + (10 +2.3) +.....+ (10 +(n-1).3) + (10 +(n+1-1).3)

    What can this be simplified to?
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  2. #2
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Its equal to the following sum.

    \sum_{k=0}^{\infty} 10 + k \cdot 3

    So simplifying this we get...

    \sum_{k=0}^{\infty} 10 + k \cdot 3 = \sum_{k=0}^{\infty} 10 + \sum_{k=0}^{\infty} k \cdot 3

    Taking each part individually...

    \sum_{k=0}^{\infty} 10 = 10n + 10

    \sum_{k=0}^{\infty} k \cdot 3 = 3\sum_{k=0}^{\infty}k = 3\sum_{k=1}^{\infty}k = 3 \frac{n(n+1)}{2}

    So the total sum is...
    <br />
10n + 10 + \frac{3n(n-1)}{2} = \frac{1}{2}(3n+20)(n+1)
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