# Binomial coefficients

• Jan 19th 2008, 02:15 PM
paolopiace
Binomial coefficients
Hello!

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

Attachment 4881

Can someone help and demonstrate it formally for me?

I tried unsuccesfully… :(

• Jan 19th 2008, 05:02 PM
topsquark
Quote:

Originally Posted by paolopiace
Hello!

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

Attachment 4881

Can someone help and demonstrate it formally for me?

I tried unsuccesfully… :(

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

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

-Dan
• Jan 19th 2008, 09:24 PM
mr fantastic
Quote:

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

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 .....

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

Apply the factorial dogmatically:

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

$\displaystyle = \frac{(-1)^k k!}{k!} = ....$
I imagine a formal approach would use the gamma function (with great care since $\displaystyle x = -n$, $\displaystyle n \in \mathbb{Z}$, 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) .....