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Homework Help: Summand part in summation notation

  1. Mar 4, 2005 #1
    I need to write the following series in summation notation

    1) 1+3+5+7+9+11 SUMMAND (2k-1)? is this right?

    2) 4+6+8+10+12+12+16+18 (2k+2)? is this right?

    Have I got it?
     
    Last edited: Mar 4, 2005
  2. jcsd
  3. Mar 4, 2005 #2

    dextercioby

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    Okay.What does these symbols mean
    [tex] \sum_{k=1}^{n} k [/tex] ?

    Daniel.
     
  4. Mar 4, 2005 #3
    [tex] \sum_{k=1}^{n} k [/tex]

    ok the n= the last number in the series

    k=1 the one is the first number in the series

    k is the summand its used to get the terms in the series u input number k through n to get the series
     
  5. Mar 4, 2005 #4
    ok for the first series i posted i got the summand to be (2k-1) with a 6 over the sigma and for the second series I got (2k+2) as the summand with 18 over the sigma, is this correct?
     
  6. Mar 4, 2005 #5

    dextercioby

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    Perfect,then u agree it means just:1+2+...+k+...+n ...?

    Okay.Now imagine how would your first sum would look like...You already did...Great.

    [tex] \sum_{k=0}^{5} (2k+1) [/tex]

    Agree...?

    Daniel.

    P.S.For some reason,we prefer the "+" for the general form of an odd #.
     
  7. Mar 4, 2005 #6

    dextercioby

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    Nope,not really.U see,the last term must coincide with the value of the general term when evaluated with the superior value:
    [tex] \sum_{k=0}^{5} (2k+1)=...+11 [/tex]

    [tex] 11=(2k+1)|_{k=5} [/tex]...

    Okay...?

    Daniel.
     
  8. Mar 4, 2005 #7
    [tex] \sum_{k=0}^{5} (2k+1) [/tex]

    ok so this is the only answer for the first series?

    [tex]\sum_{k=4}^{18} (2k+2) [/tex] and is this answer correct for the second series?
     
  9. Mar 4, 2005 #8

    dextercioby

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    Okay,true.Have your way,it's basically the same thing...:wink:

    Daniel.
     
  10. Mar 4, 2005 #9

    dextercioby

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    No,no,as i just said,your answer is true as well.Just for the first.For the second,the "k" should go from 1------>8.

    Daniel.
     
    Last edited: Mar 4, 2005
  11. Mar 4, 2005 #10
    show me how the second one looks I dont understand from 2-8?
     
  12. Mar 4, 2005 #11

    dextercioby

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    [tex] \sum_{k=1}^{8}(2k+2) [/tex] produces the same terms as the ones you had.

    Daniel.
     
    Last edited: Mar 4, 2005
  13. Mar 4, 2005 #12
    how come 2 and 8 are right the series didnt start with 2 or end at 8
     
  14. Mar 4, 2005 #13

    dextercioby

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    It's "+1" --------->"+8".It was a tiny mistake.I've edited my posts.

    [tex] (2k+2)|_{k=1}=2\times 1+2=4 [/tex]
    -----------------
    [tex] (2k+2)|_{k=8}=2\times 8+2=18 [/tex]

    Okay?

    Daniel.
     
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