∑ (–1)k Ckn = 0
(a+b)n=C0nanb0+C1nan–1b1+C2nan–2b2+...+Cknan–kbk +...+Cnnbn При a=1; b=–1 (1–1)n=C0n–C1n+C2n(–1)2+... Ckn(–1)k+...+Cnn(–1)n 0= ∑ n0(–1)kCkn