Suppose n is even, prove:
\sumk=0->n/2, C(n,2k)=2^(n-1)=\sumk=1->n/2, C(n,2k-1)
Give a combinatorial argument to prove that: (I've figured out this one...)
\sumk=1->n, C(n,k)^2=C(2n,n)
For the first problem, I tried to break C(n, 2k) into C(n+1,2k)-C(n, 2k-1), but they didnt seem to work very...