Dwumian Newtona

[thegame92]

Oblicz :

\(\sum_{k=0}^{n-1}{n \choose k}\) \(3^{k}\)

[octahedron]

\(\sum_{k=0}^{n-1}{n \choose k}3^k=\sum_{k=0}^{n}{n \choose k}3^k-3^n=\sum_{k=0}^{n}{n \choose k}3^k\cdot 1^{n-k}-3^n=(3+1)^n-3^n=4^n-3^n\)