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Coefficient of a term

  1. Mar 19, 2007 #1
    1. The problem statement, all variables and given/known data

    What is the coeff of [tex]x^{99}[/tex] in (x-1)(x-2)...(x-100)

    2. The attempt at a solution

    This has to do with the binomial coeff. I dont know how to go about it.
  2. jcsd
  3. Mar 19, 2007 #2
    Look at how a product develops as you add more terms, i.e.

    [tex](x-1)(x-2) = x^2 - (1+2)x + 2 [/tex]

    [tex](x-1)(x-2)(x-3) = x^3 -(1+2+3)x^2 + 11x + 6[/tex]

    now, what do you see?
  4. Mar 19, 2007 #3
    Look at how a product develops as you add more terms,


    [tex](x-1)(x-2)...(x-n) = x^n - (1+2+...+n)x^{n-1} + ... + (-1)^{n}*1*2*...*n [/tex]


    [tex](x-1)(x-2)...(x-n)(x-(n+1)) = x^{n+1} - (1+2+....+n+n+1)x^{n} + ... + (-1)^{n+1}*1*2*...*n*(n+1) [/tex]

    If we let n + 1 = m, then

    [tex](x-1)(x-2)...(x-m) = x^m - (1+2+...+m)x^{m-1} + ... + (-1)^{m}*1*2*...*m [/tex]
    Last edited: Mar 19, 2007
  5. Mar 19, 2007 #4


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    Homework Helper
    Gold Member

    There is an easier way.

    Here, I'll post a picture of some of hour lecture notes.

    If you understand this example, any other question like it will be a breeze.

    Attached Files:

  6. Mar 19, 2007 #5
    Hummm... I don't think it's easier for this specific problem. What we are interested in is x^n-1, which coefficient can be found by summing 1 to n and then multiplying by -1. Using combinatorics takes time, as you need to find 3 coefficients and then add them.
  7. Mar 19, 2007 #6
  8. Mar 20, 2007 #7
    Whoa. Thanks. That's really helpful. Why didnt I see that?!?
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