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Homework Help: Finding a pattern for a series and the general formula

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data

    When n = 1:
    [itex]-1 + xln(e)[/itex]

    When n = 2:
    [itex]2 - 2xln(e) + x^{2}ln(e)^{2}[/itex]

    When n = 5:
    [itex]-120 + 120xln(e) - 60x^{2}ln(e)^{2} + 20x^{3}ln(e)^{3} - 5x^{4}ln(e)^{4} + x^{5}ln(e)^{5}[/itex]

    When n = 7:
    [itex]-5040 + 5040xln(e) - 2520x^{2}ln(e)^{2} + 840x^{3}ln(e)^{3} - 210x^{4}ln(e)^{4} + 42x^{5}ln(e)^{5} - 7x^{6}ln(e)^{6} + x^{7}ln(e)^{7}[/itex]

    Identify the series and write a general formula for it (in sigma summation notation if possible).

    2. Relevant equations

    3. The attempt at a solution

    The pattern I do see is that the constant is always n!, but when n = 2, this factorial is not negative because there are an even number of terms, and the signs alternate starting from the highest power n where this term is positive.

    So the highest nth term does not have any factorials, but the subsequent terms, alternating signs, begin to increase in factorials based on what n is. For example, when n = 5, the next highest term has coefficient 5, the next term has 5*4, then 5*4*3, and finally 5*4*3*2.

    I'm not sure how to deal with the alternating signs in sigma summation notation or the factorials part.
  2. jcsd
  3. Feb 12, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This might help with a formula for the coefficients:
    5*4*3*2*1 = (5!)/(0!)
    5*4*3*2 = (5!)/(1!) (which is also equal to (5!)/(0!))
    5*4*3 = (5!)/(2!)
    5*4 = (5!)/(3!)
    5 = (5!)/(4!)
    1 = (5!)/(5!)

    Also, (-1)^n = 1 if n is even, -1 if n is odd.

    So try writing something of the form
    $$\sum_{k = 0}^{n}(\textrm{formula involving factorials})(\textrm{formula involving }(-1)^k)(\textrm{formula involving power of }x \ln(e))$$
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