Which formula satisfies the recurrence an=3an−1−2an−2a_n = 3a_{n-1} - 2a_{n-2}an=3an−1−2an−2 with initial conditions a0=1a_0 = 1a0=1 and a1=3a_1 = 3a1=3?
an=2na_n = 2^nan=2n
an=2n−1a_n = 2^n - 1an=2n−1
an=2n+1a_n = 2^n + 1an=2n+1
an=2n+1−1a_n = 2^{n+1} - 1an=2n+1−1