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Sigma Notation Question

  1. Nov 12, 2004 #1
    7
    ----
    \
    / 1 <- *note: the 1 should be more to the right.
    ----
    N=3


    Ok, this might be dumb :blushing: , but it think the answer for this is 1 but i'm just making sure.

    cuz i know that for example if it is

    7
    ----
    \
    / N <- *note: the N should be more to the right.
    ----
    N=3

    the answer would be 3+4+5+6+7

    but if it is just a 1 there, then is it just 1? i mean the N doesn't really matter because N isn't present as a variable on the right side.

    Thx guy,

    Mike.
     
  2. jcsd
  3. Nov 12, 2004 #2

    Gokul43201

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    Gold Member

    Actually, the answer is not 1.

    Here, try this : find

    [tex]\sum_{N=3}^7 (N+1)[/tex]
    and use the fact that this should be equal to [tex]\sum_{N=3}^7 N + \sum_{N=3}^7 1[/tex].

    From this you can find the last term, which is the one you want.
     
    Last edited: Nov 12, 2004
  4. Nov 13, 2004 #3
    wouldn't that be :

    (3+4+5+6+7)+(1) ??

    because

    7
    ----
    \
    / 1 <- *note: the 1 should be more to the right.
    ----
    N=3

    is = to 1 no matter what since there is no variable there?@@#%?!
     
  5. Nov 13, 2004 #4

    Gokul43201

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    When N=3, what is (N+1) = ? And what about when N= 4 ? ...
     
  6. Nov 13, 2004 #5
    ok i get your question above, but how about if it is
    7
    ----
    \
    / 1 <- *note: the 1 should be more to the right.
    ----
    N=3

    by itself? this sigma by itself would just mean 1 rite?
     
  7. Nov 13, 2004 #6
    and it doesn't have to be limited to 1 only, what about if it is 2 there, or 3, or 4..? There is no variable N there so the N shouldn't really do anything... for any value of N the value would still be the number on the right?! doesn't that make sense?
     
  8. Nov 13, 2004 #7
    ooooooooooooooooooooooooooooooooooooooook i just figured it out.

    for

    7
    ----
    \
    / 1 <- *note: the 1 should be more to the right.
    ----
    N=3

    it actually means t1 + t2 + t3 + t4

    where t1 = 1, t2 =1 and so on,

    therefore, the answer should be

    1+1+1+1 = 4 !!!!!

    is that rite Gokul43201
     
  9. Nov 13, 2004 #8

    shmoe

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    It's not 4 either, but you're closer.

    Let [tex]t_i=1[/tex] for all values of i. Your sum is

    [tex]\sum_{i=3}^{7}t_i=t_3+t_4+t_5+t_6+t_7[/tex]

    Note that you had missed a term when you rewrote it as t1+t2+t3+t4. There are actually 5, not 4, terms in this sum. If you are going to change the index variables starting point, make sure you adjust the end carefully as well.
     
  10. Nov 14, 2004 #9
    aah ok thx alot m8.
     
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