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Math Help - Newton's binomial

  1. #1
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    Newton's binomial

    I need to prepare for a math exam, however I haven't looked at anything resembling an equation in 10 years. I would appreciate any help you can provide, as I have lots of catch-up to do.

    <br />
  ^{n+3} C _{n}~-~^{n+2} C _{n-1}~=~15(n+1)<br />

    How can I find n?

    Probably I should use Newton's binomial, but I can't figure out how.
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  2. #2
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    Quote Originally Posted by avorenus View Post
    I need to prepare for a math exam, however I haven't looked at anything resembling an equation in 10 years. I would appreciate any help you can provide, as I have lots of catch-up to do.

    <br />
  ^{n+3} C _{n}~-~^{n+2} C _{n-1}~=~15(n+1)<br />

    How can I find n?

    Probably I should use Newton's binomial, but I can't figure out how.
    The first thing you need to know is that ^nC_r = {}^nC_{n-r}. One way of seeing this is to notice that choosing r objects out of n amounts to the same thing as (not) choosing the remaining n–r objects. Algebraically, both numbers are given by the same formula \frac{n!}{r!(n-r)!}.

    So ^{n+3} C _{n} - {}^{n+2} C _{n-1} =    {}^{n+3} C _{3} - {}^{n+2} C _{3}, and your equation becomes

    \frac{(n+3)(n+2)(n+1)}6 - \frac{(n+2)(n+1)n}6 = 15(n+1).

    Cancel the factor n+1 that occurs in each term, multiply out what's left, and you'll get a simple equation for n.

    The answer is
    Spoiler:
    n=28.
    Last edited by Opalg; May 29th 2009 at 04:21 AM.
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