What is the general solution for an=4an−1−3an−2a_n = 4a_{n-1} - 3a_{n-2}an=4an−1−3an−2?
an=c1(1)n+c2(3)na_n = c_1 (1)^n + c_2 (3)^nan=c1(1)n+c2(3)n
an=c1(1)n+c2(4)na_n = c_1 (1)^n + c_2 (4)^nan=c1(1)n+c2(4)n
an=c1(2)n+c2(3)na_n = c_1 (2)^n + c_2 (3)^nan=c1(2)n+c2(3)n
an=c1(−3)n+c2(1)na_n = c_1 (-3)^n + c_2 (1)^nan=c1(−3)n+c2(1)n