For the recurrence an=6an−1−9an−2a_n = 6a_{n-1} - 9a_{n-2}an=6an−1−9an−2, what is the general form of the solution?
an=c13n+c2(−3)na_n = c_1 3^n + c_2 (-3)^nan=c13n+c2(−3)n
an=(c1+c2n)3na_n = (c_1 + c_2 n) 3^nan=(c1+c2n)3n
an=(c1+c2n)(−3)na_n = (c_1 + c_2 n) (-3)^nan=(c1+c2n)(−3)n
an=c13n+c2n9na_n = c_1 3^n + c_2 n 9^nan=c13n+c2n9n