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Thread: Binomial Coefficents

  1. #1
    Junior Member
    Feb 2008

    Binomial Coefficents

    Show that (2n choose n+1) + (2n choose n) = (2n+2 choose n+1) / 2

    2n choose n+1 = 2n!/(((n+1)!)((2n-(n+1)!)) = 2n! / (((n+1)!)((n-1)!))

    2n choose n = 2n!/(n! (2n-n)!) = 2n! / (n! * n!)

    is there a way to simplify these two expressions before getting a common denominator and adding them.

    This is what I got with out simplifying and then adding them:
    [(2n! * n! * n!) + (2n! * (n+1)! (n-1)!)] / [ (n+1)! * (n-1)! * n! * n!]

    And I got stuck again trying to simplify that.
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  2. #2
    Super Member flyingsquirrel's Avatar
    Apr 2008

    To simplify, use $\displaystyle n!=n\cdot (n-1)!$ :

    $\displaystyle \binom{2n}{n+1}=\frac{(2n)!}{(n+1)!\cdot(n-1)!}=\frac{(2n)! \cdot n}{(n+1)!\cdot(n-1)!\cdot n}=\frac{(2n)!\cdot n}{(n+1)!\cdot n!}$

    $\displaystyle \binom{2n}{n}=\frac{(2n)!}{n!\cdot n!}=\frac{(2n)!\cdot (n+1)}{n!\cdot n!\cdot (n+1)}=\frac{(2n)!\cdot (n+1)}{(n+1)!\cdot n!}$

    And now it should be easier to sum. (note that you can also use this "hint" to simplify the last expression of message #1)
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