sum(C(n,k), k=0 to n) = 2^n. sum(kC(n,k), k=0 to n) = n2^(n-1). sum($k^2$*C(n,k), k=0 to n) = n(n+1)2^(n-2). Alternating: sum((-1)^kC(n,k)) = 0. These are derived by differentiating or applying operators to (1+x)^n. Appear in probability and combinatorics JEE problems.