# Binomial identity.

1. Oct 1, 2007

### MathematicalPhysicist

problem
prove that:
$$\forall n \in N \forall 0<=k<=2^{n-1} (C(2^n,k)=\sum_{j=0}^{k}C(2^{n-1},j)C(2^{n-1},k-j))$$

attempt at solution
induction seems to be too long im opting for a shorter solution, so the sum that it's wrriten in the rhs is the square of the sum of the term C(2^(n-1),j) but other than that don't know how to procceed.

2. Oct 1, 2007

### MathematicalPhysicist

C is the binomial coefficient.

3. Oct 3, 2007

### MathematicalPhysicist

so i guess no one here is that good with B.Cs?

4. Oct 4, 2007

### SanjeevGupta

Have you tried using the fact that $$2^n=2^{n-1}+2^{n-1}$$
Then expand $$(1+x)^{2^n}$$ in two different ways, equating coefficients of $$x^k$$?