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Math Help - Summation: Compute the exact sums

  1. #1
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    Summation: Compute the exact sums

    This one's a discrete mathematics question strictly due to its notation.

    Compute the exact sums of the following: (SHOW YOUR WORK)

    a) \sum_{i=k}^n7 where k is an integer between 1 and n (I have no idea how to answer this one)

    b) \sum_{k=1}^n(3k^2-2n+8) (I believe I have the answer, but I lack the steps and process to reach the answer)

    c) \sum_{j=1}^n\sum_{k=1}^j\sum_{i=k}^n(i-k) (I may have answered this one, but I'd like confirmation)

    d) \sum_{j=18}^{99}12 (This one was easy, and I've already answered it)

    This shouldn't be too hard for some people, but I'd rather not get these wrong.
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  2. #2
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    Quote Originally Posted by Runty View Post
    a) \sum_{i=k}^n7 where k is an integer between 1 and n (I have no idea how to answer this one)
    1 \leqslant k \leqslant n \Rightarrow \quad \sum\limits_{i = k}^n 7  = 7\left( {n - k + 1} \right)
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  3. #3
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    Hi runty,
    for 2 you need to rewrite the sum and isolate the term which is being summed:

    <br />
\sum_{k = 1}^n (3k^2 - 2n + 8) = (\sum_{k = 1}^n 3k^2) - n(2n + 8) = 3(\sum_{k = 1}^n k^2) - n(2n + 8)<br />

    Check this! Do you need n(2n + 8)?
    Then you can use a closed form of the sum of squares to replace the sum with a relation based on n:

    <br />
\sum_{k = 1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}<br />
    Square pyramidal number - Wikipedia, the free encyclopedia
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  4. #4
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    b)

    \sum_{k=1}^{n} 3k^2-2n+8 = 3\sum_{k=1}^{n}k^2 -2\sum_{k=1}^{n}n + \sum_{k=1}^{n}8

    \frac{n(n+1)(2n+1)}{2} - 2n^2 + 8n
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