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Help! What are the patterns

  1. Nov 1, 2005 #1
    I have a tricky question here....

    Please just feed in any comments, anything you can spot that I can't

    I have been asked to compare the formulae of Sygma (k=1 until n value) of k, Sygma (k=1 until n value) of k(k+1) and Sygma (k=1 until n value) of k(k+1)(k+2)?

    Hence comparing these...

    Sygma (k=1 until n value) of k = 1/2 n(n+1)
    Sygma (k=1 until n value) of k(k+1) = 1/3 n(n+1)(n+2)
    Sygma (k=1 until n value) of k(k+1)(k+2) = 1/4 n(n+1)(n+2)(n+3)

    I have been asked to comment on these as much as I possibly can??? :confused: Can anyone help?
     
    Last edited: Nov 1, 2005
  2. jcsd
  3. Nov 1, 2005 #2
    poo, no one has a clue :uhh:
     
  4. Nov 2, 2005 #3

    benorin

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    Homework Helper

    Here you go, Natasha.

    Here is a helpful observation: [tex]k(k+1)\cdot\cdot\cdot(k+j)=\frac{(k+j)!}{(k-1)!}[/tex] and [tex]\sum_{k=1}^{n}\frac{(k+j)!}{(k-1)!}=\frac{n(n+j+1)!}{(j+2)n!}[/tex]
    Note that [tex]k[/tex], [tex]k(k+1)[/tex], and [tex]k(k+1)(k+2)[/tex] are given by [tex]\frac{(k+j)!}{(k-1)!}[/tex] for [tex]j=0,1,\mbox{ and }2,[/tex] respectively, so that

    [tex]\sum_{k=1}^{n} k = \frac{n(n+0+1)!}{(0+2)n!} = \frac{n(n+1)}{2}[/tex],

    [tex]\sum_{k=1}^{n} k(k+1) = \frac{n(n+1+1)!}{(1+2)n!} = \frac{n(n+1)(n+2)}{3}[/tex],

    and

    [tex]\sum_{k=1}^{n} k(k+1)(k+2) = \frac{n(n+2+1)!}{(2+2)n!} = \frac{n(n+1)(n+2)(n+3)}{4}[/tex].
    -Ben
     
    Last edited: Nov 2, 2005
  5. Nov 2, 2005 #4

    Tide

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    Ben,

    That is REALLY nice!
     
  6. Nov 2, 2005 #5
    Can anyone see anything else please? :uhh:
     
  7. Nov 5, 2005 #6
    thanks super Ben!
     
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