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Math Help - binomial expansion

  1. #1
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    binomial expansion

    Can someone please help me with this question

    Use the inductive property:
    nC(r+1) = (n-r)/(r+1) * nCr

    to prove the Pascal triangle property that nCr + nC(r+1) = (n+1)C(r+1)


    This is how I started..
    nCr + (n-r)/(r+1) * nCr = (n+1-r)/(r+1) * nCr -->
    nCr (1+ (n-r)/(r+1)) = nCr (n+1-r)/(r+1)

    I don't know what to do next..
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  2. #2
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    Re: binomial expansion

    Hey puresoul.

    Hint: (r+1)*r! = (r+1)!
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  3. #3
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    Re: binomial expansion

    Quote Originally Posted by chiro View Post
    Hey puresoul.

    Hint: (r+1)*r! = (r+1)!
    Thnks for your hint..

    I took another approach..

    n!/(r!(n-r)!) + n!/ ( (r+1)!(n-(r+1))!

    Then i tried to multiply the first fraction by (r+1) so i got in the denominator (r+1)!, but i didn't know how to simplify the top part and to get (n+1)!
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  4. #4
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    Re: binomial expansion

    I'm going to attempt it straight from the Pascal relationship:

    nCr + nC(r+1)
    = n!/[r!*(n-r)!] + n!/[(r+1)!*(n-r-1)!]
    = n!/[r!*(n-r-1)!] * [1/(n-r) + 1/(r+1)]
    = n!/[r!*(n-r-1)!] * [r+1 + n - r]/[(n-r)(r+1)]
    = n!/[r!*(n-r-1)!] * [n+1]/[(n-r)(r+1)]
    = (n+1)!/[(r+1)!*(n-r)!]
    = (n+1)C(r+1)
    Thanks from puresoul
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  5. #5
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    Re: binomial expansion

    Quote Originally Posted by chiro View Post
    I'm going to attempt it straight from the Pascal relationship:

    nCr + nC(r+1)
    = n!/[r!*(n-r)!] + n!/[(r+1)!*(n-r-1)!]
    = n!/[r!*(n-r-1)!] * [1/(n-r) + 1/(r+1)]
    = n!/[r!*(n-r-1)!] * [r+1 + n - r]/[(n-r)(r+1)]
    = n!/[r!*(n-r-1)!] * [n+1]/[(n-r)(r+1)]
    = (n+1)!/[(r+1)!*(n-r)!]
    = (n+1)C(r+1)
    THNKS A LOT ^_____^
    I got it !!
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