Suppose n is even, prove: [tex]\sum[/tex]k=0->n/2, C(n,2k)=2^(n-1)=[tex]\sum[/tex]k=1->n/2, C(n,2k-1) Give a combinatorial argument to prove that: (I've figured out this one...) [tex]\sum[/tex]k=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 well. (I also noticed that [tex]\sum[/tex]0, n C(n,2k)=2^n, but I can't really use this to solve the problem.) For the second problem, I don't have a clue. Any thoughts? Thanks!