Solve the recurrence an=3an−1a_n = 3a_{n-1}an=3an−1 with a0=1a_0 = 1a0=1.
an=3na_n = 3^nan=3n
an=3na_n = 3nan=3n
an=3n−1a_n = 3^{n-1}an=3n−1
an=n3a_n = n^3an=n3