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Sigma notaion help

  1. Sep 30, 2007 #1
    can some one show me how to write
    5+10+17+26 in sigma notation
     
  2. jcsd
  3. Sep 30, 2007 #2

    rock.freak667

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    do you notice anything about those numbers? From 5 how do you get to 10? from 10 how to get to 17? from 17 how to get to 26? etc
     
  4. Sep 30, 2007 #3

    CRGreathouse

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    One way would be
    [tex]S=\{5,10,17,26\}[/tex]
    [tex]\sum_{x\in S}x[/tex]

    Another:
    [tex]s_1=5,s_2=10,s_3=17,s_4=26[/tex]
    [tex]\sum_{k=1}^4s_k[/tex]

    Another:
    [tex]\sum_{k=1}^4k^2+2k+2[/tex]
     
  5. Oct 1, 2007 #4

    cks

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    I can get the nth term that is

    Tn=2+[tex]\sum_{k=1}^n2k+1[/tex]

    but I don't know how to get from the Tn to Sn.
     
    Last edited: Oct 1, 2007
  6. Oct 1, 2007 #5

    cks

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    How CRGreatHouse see that or how did he trasform from the nth term to the Sn term?
     
  7. Oct 1, 2007 #6

    cks

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    [tex]\S_n=sum_{b=1}^nT_b[/tex]

    where

    [tex]\T_b=2+sum_{k=1}^b2k+1[/tex]
     
  8. Oct 1, 2007 #7

    cks

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    this is what I can do at most!
     
  9. Oct 1, 2007 #8

    matt grime

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    There is no 'unique' or 'canonical' way to do it. There are many ways. Yours (it appears to me) is fine.
     
  10. Oct 1, 2007 #9

    CRGreathouse

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    I noticed that the second differences were constant, which means that a quadratic can be uniquely fitted to it. The sequence is 5, 10, 17, 26; the first differences are 5, 7, 9; the second differences are 2, 2. I then solved the system a+b+c=5, 4a+2b+c=10, 9a+3b+c=17. (I could have used 16a+4b+c=26 but didn't need it.)
     
    Last edited: Oct 2, 2007
  11. Oct 1, 2007 #10

    cks

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    I try some examples and see that when there's a difference of 2, then we can somewhat say that the nth term contains k^2,

    well, but i fail to guess how you come out with the linear system, a+b+c=5, 4a+2b+c=10, 9a+3b+c=17.

    Sorry, I have been trying hard to think about it.
     
  12. Oct 2, 2007 #11

    CRGreathouse

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    I'm looking for a solution to y = ax^2 + bx + c with (x, y) = (1, 5), (2, 10), and (3, 17). Substitute and you get the linear system above.
     
  13. Oct 2, 2007 #12

    HallsofIvy

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    In general, you can use "Newton's Difference Formula":
    If you have a sequence {an} so that a0= a, the first difference, a1- a0, is b, the "second difference" (subtract the first two first differences) is c, etc. then an= a+ bn+ (c/2)n(n-1)+ ... It looks a lot like a Taylor's series formula but you use n(n-1)(n-2)...(n-k) instead of xk.
     
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