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Homework Help: Series, Sequence and Probablility Question

  1. Sep 28, 2005 #1
    I am working with problems which are taking a toll on me.
    1st and 3rd partial sums of the sequence An=(-2)^n+5

    ----I don't even know what formula to use to start this problem

    and sum of hte geometric series:
    2/3 - 4/9 + 8/27 - ...

    I think I use this formula for this one: S=A1/1-r
    Please help :eek:
  2. jcsd
  3. Sep 28, 2005 #2


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    Do you understand what a "partial sum" is? That first problem is just asking you to find A1= (-2)1+ 5 (the "first partial sum") and then
    A1+ A2+ A3= ((-2)1+5)+ ((-2)2+ 5)+ ((-2)3+ 5).

    Yes, the sum of an infinite sum a+ ar+ ar2+ ... is a/(1-r).
    Here, you have (2/3)+ (2/3)(-2/3)+ (2/3)(-2/3)2+... What are a and r?
  4. Sep 28, 2005 #3
    Thanks for the help, I have come up with S1=3 and S3=-3

    For the second I have come up with an answer of .518

    Am I correct?
  5. Sep 28, 2005 #4


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    Your S1 is correct, but I think you should re-check your answer for S3, S3 = A1 + A2 + A3.
    And also, how did you come up with .518 in #2???
    Your first term is 2 / 3. And all you need to do is to find r. So what do you get for r?
    Viet Dao,
  6. Sep 29, 2005 #5
    For the S3= -3
    ((-2)3+ 5) = -3 (Am I not doing this correctly?)

    #2:For r, I have r=1/3
  7. Sep 29, 2005 #6
    Can anyone lend some advice?
  8. Sep 29, 2005 #7


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    Nope, you are not doing it correctly.
    Sn is the sum of the first n terms.
    So S3 is the sum of the first 3 terms. So:
    S3 = A1 + A2 + A3 = ...
    Note that they are not asking for A3, they are asking for S3.
    So what do you get for S3? :smile:
    How can you come up with r = 1 / 3???
    [tex]a_1 = \frac{2}{3}[/tex]
    [tex]a_2 = -\frac{4}{9} = a_1r[/tex]
    So again, what is r?
    Viet Dao,
    Last edited: Sep 29, 2005
  9. Sep 29, 2005 #8
    Ah ha! I have come up with 9 for S3.

    I am still baffled with the second. I am not grasping something.
  10. Sep 29, 2005 #9


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    Your sum is 2/3 - 4/9 + 8/27 -....

    The "general" geometric series is a+ ar+ ar2+ ar3+...

    Obviously "a" is just the first term: 2/3. r= ar/r is just the second term divided by the first term: -(4/9)/(2/3)= what?

    Now put those into a/(1-r)
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