# Thread: Proof for an alternative binomial coefficient

1. ## Proof for an alternative binomial coefficient

Could anyone please point me in the direction on how to proof $C(-n, k) = (-1)^k C(n + k - 1, k)$ ?

I'm having a problem with this as the factorial of a negative integer is infinit.

2. Originally Posted by chris2547
Could anyone please point me in the direction on how to proof $C(-n, k) = (-1)^k C(n + k - 1, k)$ ?

I'm having a problem with this as the factorial of a negative integer is infinit.

You mean $C(n,k)$ instead of $C(-n,k)$

Beside $(-1)^k$ is not infinity.

3. Originally Posted by novice
You mean $C(n,k)$ instead of $C(-n,k)$

Beside $(-1)^k$ is not infinity.
No, I'm pretty sure I mean $C(-n,k)$; and did you ever try (-n!) ?

4. Originally Posted by chris2547
No, I'm pretty sure I mean $C(-n,k)$; and did you ever try (-n!) ?
I assume by $C(-n,k)$ you mean the coefficient of $x^k$ in the binomial expansion of $(1+x)^{-n}$

CB

5. Originally Posted by chris2547
Could anyone please point me in the direction on how to proof $C(-n, k) = (-1)^k C(n + k - 1, k)$ ?

I'm having a problem with this as the factorial of a negative integer is infinit.

The starting point in this is to generalize the definition of C(x, k) to be

$C(x,k) = \frac{x (x-1) (x-2) \cdots (x-k+1)}{k!}$.

With this definition, x can be any real number, not necessarily an integer. k must be a non-negative integer.

Now let x = -n and see what you get.

6. Originally Posted by awkward
The starting point in this is to generalize the definition of C(x, k) to be

$C(x,k) = \frac{x (x-1) (x-2) \cdots (x-k+1)}{k!}$.

With this definition, x can be any real number, not necessarily an integer. k must be a non-negative integer.

Now let x = -n and see what you get.
Thanks alot! I got it now. I just missed that one detail about the generalization of the binomial coëfficiënt.