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What will the following formula go ?

  1. May 20, 2004 #1
    Can you point out how to simplify the following formula ?


    Thank you
  2. jcsd
  3. May 20, 2004 #2

    matt grime

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    your Cs are binomial coeffs?

    have you worked itout for the cases n=0,1,2? what did you get there? is there a pattern you can see?
  4. May 20, 2004 #3
    Thanks Matt for your help

    Yes, they are binomial coeffs.
    I tried till n reaches 4 and I figured it out...
    [tex]n=0 \frac{C^0_0}{x}=\frac{1}{x}[/tex]
    [tex]n=1 \frac{C^0_1}{x}-\frac{C^1_1}{x+1}=\frac{1}{x(x+1)}[/tex]
    [tex]n=2, \frac{C^0_2}{x}-\frac{C^1_2}{x+1}+\frac{C^2_2}{x+2}=\frac{1}{x}-\frac{2}{x+1}+\frac{1}{x+2}=\frac{2}{x(x+1)(x+2)}[/tex]
    [tex]n=3, \frac{C^0_3}{x}-\frac{C^1_3}{x+1}+\frac{C^2_3}{x+2}-\frac{C^3_3}{x+3}=\frac{6}{x(x+1)(x+2)(x+3)}[/tex]
    [tex]n=4, \frac{C^0_4}{x}-\frac{C^1_4}{x+1}+\frac{C^2_4}{x+2}-\frac{C^3_4}{x+4}+\frac{C^4_4}{x+4}=\frac{24}{x(x+1)(x+2)(x+3)(x+4)}[/tex]

    So I think it will be [tex]\frac{n!}{x(x+1).....(x+n)}[/tex]

    Thank Matt so very much for your suggestions, :smile:
    Last edited: May 20, 2004
  5. May 20, 2004 #4

    matt grime

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    that seems about right:

    let P=x(x+1)(x+2)...(x+n) and let P(r) be P, but where you omit the factor (x+r)

    then you want to work out
    {P(0) -P(1)nC1 + P(2)nC2 ....)/P

    if you work out the coeff of x^s in the bracket you see lots of things happening:

    x^(n-1) has coeff the alternating sum of all the binom coeffs, so it's zero,
    x^0 is just the constant term in P(0), cos all the other terms P(s) have a factor of x in them. you should tidy up that to work for all coeffs
  6. May 20, 2004 #5
    Oh Well, That is really great, I have just learnt new things from you, Matt. :sm:

    Thank Matt very much...
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