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Homework Help: Find coefficient on power series

  1. Mar 1, 2010 #1
    (Moderator's note: thread moved from "General Math")

    I am confused with this question. I tried two different ways to solve it, but I got different answers for each way. The question is
    "Determine the coefficient of x^100 in the pwer series form of (1+x+x^2)/((1-x^3)^2)"
    First, I tried this way,
    since we know 1/(1-x^3)=(1+x^3+x^6+x^9+.....), so the form becomes
    let a0,a1 be the power of x from each series,then a0+a1=100.
    from the first part we know that a0=0 or 1 or 2.
    When a0=0, a1^2=a1+a1=100
    when a0=1, a1^2=a1+a1=99
    when a0=2, a1^2=a1+a1=98
    and we know that a1 is multiple of 3, let a1=3n, then 6n=100.
    Since we do not have n value, the coefficient of x^100 is 0.
    Second I tried this way,
    since we know 1+x+x^2=(1-x^3)/(1-x), if we apply this to our form, (1-x^3) from top and bottom is gone, therefore we have (1/(1-x))*(1/(1-x^3)).
    since we know, 1/(1-x)=1+x+x^2+x^3+....
    and 1/(1-x^3)=1+x^3+x^6+x^9+...
    therefore we now have (1+x+x^2+x^3+...)*(1+x^3+x^6+x^9+...)
    let a0 and a1 be the power of x from each series, then a0+a1=100.
    Since a1 is the multiple of 3, it can be written as 3a1, then a0+3a1=100.
    We have (a0,a1) as (1,33),(10,30),(19,27),(28,24)...(100,0) there are 12 sets
    and each has coefficient of 1,
    therefore the coefficient is 12.
    I got two different answers. I don't know what is the problem...
    Please tell me whether I am right or wrong...
    Last edited by a moderator: Mar 3, 2010
  2. jcsd
  3. Mar 2, 2010 #2


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    Science Advisor

    Both calculations are incorrect!!
    For the first approach when a0=1, the following a1 pairs are allowed (0,99), (3,96),...(96,3), (99,0). 34 pairs.
    For the second set, you need the following (a0,a1) pairs (1,99), (4,96),....(97,3), (100,0). 34 pairs.
    Last edited: Mar 2, 2010
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