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Math Help - Sum The Series

  1. #1
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    Arrow Sum The Series

    How would I solve this?:

    Sum The Series-

    1 x 2 + 2 x 3 + 3 x 4 + . . . + 1994 x 1995

    Thanx
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  2. #2
    MHF Contributor red_dog's Avatar
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    \displaystyle\sum_{k=1}^{1994}k(k+1)=\sum_{k=1}^{1  994}(k^2+k)=\sum_{k=1}^{1994}k^2+\sum_{k=1}^{1994}  k
    Now use the formulas
    \displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6  } and \displaystyle\sum_{k=1}^nk=\frac{n(n+1)}{2}
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  3. #3
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    Hello, red_dog!

    Great solution!


    \displaystyle\sum_{k=1}^{1994}k(k+1) \;= \;\sum_{k=1}^{1994}(k^2+k)\;=\;\sum_{k=1}^{1994}k^  2+\sum_{k=1}^{1994}k

    Now use the formulas:

    \displaystyle\sum_{k=1}^nk^2\;=\;\frac{n(n+1)(2n+1  )}{6} .and . \displaystyle\sum_{k=1}^nk\;=\;\frac{n(n+1)}{2}
    I took it a step further . . .

    \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \;=\; \frac{n(n+1)(2n + 1 + 3)}{6} \;=\;\frac{n(n+1)(n+2)}{3}

    A curious result: the product of three consecutive integers divided by 3.

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  4. #4
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    This result can be generalized to n consecutive sums of products.
    (Likewise, with reciprocals too).
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