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The sum of an infinite geometric series

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data

    1+(x+1)+(x+1)^2+(x+1)^3 + ..... if lx+1l < 1



    2. Relevant equations

    Sn=a/1-r



    3. The attempt at a solution

    My attempt:

    so I have a = x+1 and r = x+1
    from there i get x+1/1-(x+1)
    which is x+1/1-x-1
    from there x+1/-x
    multiply by the reciprocal

    my solution is x^2 + x + 1 with the 1 coming from the original sequence.

    What do you think?
     
  2. jcsd
  3. Oct 20, 2008 #2
    I think the one should go and the answer is: -(x+1)/x (can you see why?)
     
  4. Oct 20, 2008 #3
    So do I have everything correct until

    from there x+1/-x?

    If so then yes I see where you got it from.

    x+1/-x the reciprocal is

    x+1 multiplied by -1/x

    correct?
     
  5. Oct 20, 2008 #4
    Also, for this sequence, would 1 be the a variable? I have x+1 as the a variable but why wouldnt it be 1?
     
  6. Oct 21, 2008 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The first term in the series is 1. What does that tell you?
     
  7. Oct 21, 2008 #6
    If 1 is the first term, then everything leading up to -(x+1)/x is wrong. So what is the first term?

    If 1 is the first term then the answer is 1/-x? Can I leave the answer like that?
     
    Last edited: Oct 21, 2008
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