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Math Help - Closed form of sum

  1. #1
    Member Mollier's Avatar
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    Closed form of sum

    Hi,

    I was just playing around end ended up trying to find a closed form of the following sum,
    <br />
\sum^{n-1}_{i=1} i(n-i)

    Writing it out I get,

    1(n-1)+2(n-2)+3(n-3)+\cdots+ (n-1)(1)

    which is,

    2(n-1)+2(n-2)+3(n-3)\cdots+ (n-2)(2)

    which is,

    2(n-1)+4(n-2)+3(n-3)\cdots (n-3)(3) and so on...

    I'm not sure what to make of this though. Could someone help me out finding a closed form for this sum?
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  2. #2
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    This can be calculated using the formula for the sum of numbers from 1 to n (arithmetic progression) and the sum of squares. (There is also a formula for the sum of cubes -- the first that popped up in a search.)
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  3. #3
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by emakarov View Post
    This can be calculated using the formula for the sum of numbers from 1 to n (arithmetic progression) and the sum of squares. (There is also a formula for the sum of cubes -- the first that popped up in a search.)
    Just to elaborate a bit (maybe you don't need it, but could be useful for others looking at the thread),

    \displaystyle \sum^{n-1}_{i=1} i(n-i)=\sum^{n-1}_{i=1} (in-i^2) = n\sum^{n-1}_{i=1}i - \sum^{n-1}_{i=1} i^2
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  4. #4
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    Hello, Mollier!

    I was just playing around end ended up trying to find

    a closed form of the following sum: . \displaystyle \sum^{n-1}_{i=1} i(n-i)


    \displaystyle n=2\!:\;\sum^1_{i=1}i(2-i) \;=\;1\cdot1 \:=\:1

    \displaystyle n = 3\!:\;\sum^2_{i=1}i(3-i) \;=\;1\cdot2 + 2\cdot1 \;=\;4

    \displaystyle n-4\!:\;\sum^3_{i=1}i(4-i) \;=\;1\cdot3 + 2\cdot2 + 3\cdot1 \;=\;10

    \displaystyle n-5:\;\sum^4_{i=1}i(5-i) \;=\;1\cdot4 + 2\cdot3 + 3\cdot2 + 4\cdot1 \;=\;20

    \displaystyle n = 6\!:\;\sum^5_{i=1}i(6-i) \;=\;1\cdot5 + 2\cdot4 + 3\cdot3 + 4\cdot2 + 5\cdot1 \;=\;35

    \displaystyle n=7\!:\;\sum^6_{i=1}i(7-i) \;=\;1\cdot6+ 2\cdot5+ 3\cdot4 + 4\cdot3 + 5\cdot2 + 6\cdot1 \;=\;56



    The sequence: . 1,\:4,\:10,\:20,\:35,\:56,\:\hdots are Tetrahedral Numbers.


    They are the number of balls in a triangular pyramid of balls.
    You must imagine these triangles "stacked up".

    n = 2\qquad \begin{array}{c}\circ \end{array}


    n = 3 \qquad \begin{array}{c} \circ\end{array} \quad \begin{array}{c} \circ \\ [-3mm] \circ \circ \end{array}


    n = 4 \qquad \begin{array}{c}\circ\end{array} \quad \begin{array}{c}\circ \\ [-3mm] \circ\circ\end{array} \quad \begin{array}{c}\circ \\ [-3mm] \circ\circ \\ [-3mm]\circ\!\circ\!\circ\end{array}

    n = 5\qquad\begin{array}{c}\circ\end{array} \quad \begin{array}{c}\circ \\ [-3mm] \circ\circ\end{array} \quad \begin{array}{c}\circ \\ [-3mm] \circ\circ \\ [-3mm]\circ\!\circ\!\circ\end{array} \quad \begin{array}{c}\circ \\ [-3mm] \circ\circ \\ [-3mm] \circ\!\circ\!\circ \\ [-3mm] <br />
\circ\!\circ\!\circ\circ \end{array}


    The n^{th} term is this sequence is: . a_n \:=\:\dfrac{(n-1)n(n+1)}{6}
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  5. #5
    Member Mollier's Avatar
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    That is just wonderful guys, thank you!
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