Chứng minh: C 0 n 3^n + C 1 n #^(n - 1) + + C k n 3^(n - k) + + C (n - 1) n 3 + C n n
Giải thích
Cn03n+Cn13n−1+...+Cnk3n−k+...+Cnn−13+Cnn
=Cn03n+Cn13n−1.11+...+Cnk3n−k.1k+...+Cnn−13.1n−1+Cnn.1n
=3+1n=4n.
Cn0+Cn13+...+Cnk3k+...+Cnn−13n−1+Cnn3n
=Cn0.1n+Cn1.1n−13+...+Cnk.1n−k3k+...+Cnn−1.1n−13n−1+Cnn3n
=1+3n=4n.
Vậy Cn03n+Cn13n−1+...+Cnk3n−k+...+Cnn−13+Cnn
=Cn0+Cn13+...+Cnk3k+...+Cnn−13n−1+Cnn3n.