A recurrence relation an=3an−1−2an−2a_n = 3a_{n-1} - 2a_{n-2}an=3an−1−2an−2 has the general solution an=c1(1n)+c2(2n)a_n = c_1(1^n) + c_2(2^n)an=c1(1n)+c2(2n). What is the solution if a0=1,a1=3a_0=1, a_1=3a0=1,a1=3?
an=2n−1a_n = 2^n - 1an=2n−1
an=2n+1−1a_n = 2^{n+1} - 1an=2n+1−1
an=2n+1a_n = 2^n + 1an=2n+1
an=2n+1+1a_n = 2^{n+1} + 1an=2n+1+1