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Math Help - Explicit form

  1. #1
    Senior Member
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    Explicit form

    1*2+2*3+3*4+\cdots +(n-1)n=?
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  2. #2
    Junior Member
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    1\times2+2\times3+3\times4+\ldots+(n-1)n
    =2+6+12+20+\ldots+(n^2-n)
    =2\left(1+3+6+10+\ldots+\frac{n^2-n}{2}\right)
    Inside the parenthesis is the sum of the first n-1 triangular numbers. This can be given by the formula,
    \frac{(n-1)(n)(n+1)}{6}
    Then,
    2\left(1+3+6+10+\ldots+\frac{n^2-n}{2}\right)<br />
=\frac{(n-1)(n)(n+1)}{3}
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  3. #3
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    Hello, James!

    Find the explicit form for: . 1\cdot2+2\cdot3+3\cdot4 + \cdots +(n-1)n
    It helps if you know some summation formulas . . .


    We have: . \sum^n_{k=1}k(k-1) \;=\;\sum^n_{k=1}(k^2 - k)

    . . =\qquad\quad\sum^n_{k=1} k^2 \quad-\qquad \sum^n_{k=1} k

    . .  =\;\overbrace{\frac{k(k+1)(2k+1)}{6}} - \overbrace{\frac{k(k+1)}{2}}


    Factor: . \frac{k(k+1)}{6}\cdot\left[(2k+1) - 3\right] \;=\;\frac{k(k+1)}{6}\cdot(2k-2) \;=\;\frac{k(k+1)}{6}\cdot2(k-1)


    Therefore: . \boxed{\frac{(n-1)n(n+1)}{3}}

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