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Math Help - Binomial coefficients

  1. #1
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    Binomial coefficients

    Hello!

    Working on the binomial series I found (empirically) this equality:

    Binomial coefficients-screenshot010.bmp

    Can someone help and demonstrate it formally for me?

    I tried unsuccesfully…

    Thanks in advance!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by paolopiace View Post
    Hello!

    Working on the binomial series I found (empirically) this equality:

    Click image for larger version. 

Name:	ScreenShot010.bmp 
Views:	40 
Size:	23.6 KB 
ID:	4881

    Can someone help and demonstrate it formally for me?

    I tried unsuccesfully…

    Thanks in advance!
    Okay, I'll bite. Is this supposed to be the combinitorial function? Typically
    {n \choose r} = \frac{n!}{r!(n - r)!}

    How can you define this for a negative n? Or does your symbol mean something else?

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    Okay, I'll bite. Is this supposed to be the combinitorial function? Typically
    {n \choose r} = \frac{n!}{r!(n - r)!}

    How can you define this for a negative n? Or does your symbol mean something else?

    -Dan
    You can do it in a sloppy way .....

    {-1 \choose k} = \frac{(-1)!}{k!(-1 - r)!}

    Apply the factorial dogmatically:

    = \frac{(-1)(-2)(-3) .... (-k)(-k-1)!}{(-1 - k)! k!}

    Cancel the bad stuff:

    = \frac{(-1)^k k!}{k!} = ....

    And you're left with the good stuff.


    I imagine a formal approach would use the gamma function (with great care since x = -n, n \in \mathbb{Z}, are simple poles .....) Maybe if someone has the time and the inclination (but I don't think a pre-algebra and algebra forum is the place) .....
    Last edited by mr fantastic; January 19th 2008 at 09:36 PM.
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    To mr fantastic...

    That's the way I did prior to posting here.

    I'd really like a more formal proof...
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