No clue yet?
Thank you for your comment.
You expressed 2^n as sum of combinations. I know the expansion.
My goal is different.
I tried to find how to express C(n,k) as summation.
Here is what I mean.
If we call k the degree of C(n,k) I have found an identity such as :
C(n,k)= sum (a(i)*C(b(i),k-2)) i varying from b to d
I have found it while working on C(n,5).
I tried to google without success.
We could expand any C(n,k) by recurrence as sum of products.
C(n,k) .....> C(m,k-2)
C(m,k-2) .....> C(l,k-4)
and so on
I checked it for C(n,5).
I`m aware that such result is not useful practically.
But it could help proving other identities or equations.
Hi,
As I do not know how to use Latex here is the formula I discovered recently (look at the attachment).
What I want to know :
Is the formula correct?
How to prove it?
Is the formula known?
What can we do with such formula?
Thank you for your help