Recent content by Puzzedd

Thanks! I spent an hour thinking about it last night and couldn't see what I was doing wrong.

Using summation((\stackrel{n}{k})xkyn-k) = (x+y)n, I let x = y = 1. This should then result in summation((\stackrel{n}{k})*1*1) = (1 + 1)n = 2n. Expanding the summation, I get (\stackrel{n}{0}) + (\stackrel{n}{1}) + ... +(\stackrel{n}{n}) = 2n. Solving this results in...