A recurrence relation an=3an−1−2an−2a_n = 3a_{n-1} - 2a_{n-2}an=3an−1−2an−2 has roots r1=1,r2=2r_1=1, r_2=2r1=1,r2=2. What is the general solution?
an=c1(1)n+c2(2)na_n = c_1(1)^n + c_2(2)^nan=c1(1)n+c2(2)n
an=c1(2)n+c2(1)na_n = c_1(2)^n + c_2(1)^nan=c1(2)n+c2(1)n
an=c1+c22na_n = c_1 + c_2 2^nan=c1+c22n
All of the above