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Finding the nth term,,

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data

    This type of questions seems to be easy, but a little confusing at the same time
    If we are given the Sn of a series, and were asked to find an how to do that ?


    2. Relevant equations

    the limit of series and Sn is the same

    3. The attempt at a solution

    Maybe we should divide an and Sn ??
     
  2. jcsd
  3. Apr 4, 2009 #2

    Dick

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    What's the difference between S_n and S_(n+1)?
     
  4. Apr 4, 2009 #3
    I think that the only difference is one comes before the other, however they both have the same limits
     
  5. Apr 4, 2009 #4

    Dick

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    Maybe you'd better define what S_n means. I thought I knew what you were talking about. But your answer isn't reassuring.
     
  6. Apr 4, 2009 #5
    Sn is the partial sum of the nth term of a series .
     
  7. Apr 4, 2009 #6

    Dick

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    No, it's the partial sum of the first n terms in the series. a1+a2+a3+...+an. What would 'partial sum of an nth term' mean?
     
  8. Apr 4, 2009 #7
    Uha, and how to use to find an ?? When we know Sn as stated in the question ?
     
  9. Apr 4, 2009 #8

    Dick

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    I'll ask you again. Using slightly different words. What's S_(n+1)-S_(n)?
     
  10. Apr 4, 2009 #9
    I think that this is an right ?

    SO do I have to find S(n+1) and subt. from S (n) ??

    Is that what you mean ???
     
  11. Apr 4, 2009 #10

    Dick

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    That was supposed to be a hint, not the answer. I would say S_(n+1)-S_(n)=a_(n+1). You'll have to change it a little to get a_(n). But, yes, that's what I mean.
     
  12. Apr 4, 2009 #11
    mmm, ok what do you think of this
    If the Sn = 3-n(2)^(-n )

    I did what you suggested and I got that an = 2^(-n) ( n+2n+2)

    Do you think that can be right ???
     
  13. Apr 4, 2009 #12

    Dick

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    That's not what I get. How did you work that out?
     
  14. Apr 4, 2009 #13
    an = Sn - S (n-1)

    I found that Sn-1 = 3-(n-1)2^(-n+1)

    Is that right, before I precede ???
     
  15. Apr 4, 2009 #14

    Dick

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    Looks ok to me.
     
  16. Apr 4, 2009 #15
    Ya, and that 's how I got the rest !!

    I think I am fine "with this one "

    Thanks alot Dick
     
  17. Apr 4, 2009 #16

    Dick

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    Ok, you're welcome! But I still don't get 2^(-n) ( n+2n+2).
     
  18. Apr 4, 2009 #17
    I got that by sub. Sn - S(n-1)

    and then I took 2^(-n) as a common factor..

    What do you think ?
     
  19. Apr 4, 2009 #18

    Dick

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    I think it's wrong. If you'd show the rest of your work maybe we could figure out why.
     
  20. Apr 4, 2009 #19
    Uha, sure :

    here what I did :
    Sn- S(n-1) = 3-n2(-n) - [ 3- n 2^(-n+1) - 2^ (-n+1) ]
    3 we cancel and then we got what told you defore a while ,,

    What do you think ?
     
  21. Apr 4, 2009 #20

    Dick

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    The '3's cancel, sure. I'm left with (n-1)*2^(-n+1)-n*2^(-n). If I factor 2^(-n) out I've got 2^(-n)*((n-1)*2-n). Right? Be more careful with the signs.
     
    Last edited: Apr 4, 2009
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