Hoặc
Cho n ∈ ℕ*. Chứng minh Cn0+Cn1+Cn2+...+Cnn−1+Cnn=2n .
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=2n .