Cho n thuộc N*. Chứng minh C 0 n + C 1 n + C 2 n + + C (n -1) n + C n n = 2^n

Ta có:

x+1n=Cn0xn+Cn1xn−1 . 1+Cn2xn−2 . 12+...+Cnn−1x . 1n−1+Cnn . 1n=Cn0xn+Cn1xn−1+Cn2xn−2+...+Cnn−1x +Cnn.

Cho x = 1, ta được:

1+1n=Cn01n+Cn11n−1+Cn21n−2+...+Cnn−11 +Cnn=Cn0+Cn1+Cn2+…+Cnn−1+Cnn.

Vậy Cn0+Cn1+Cn2+…+Cnn−1+Cnn=1+1n=2n.