Given the recurrence relation an=6an−1−9an−2a_n = 6a_{n-1} - 9a_{n-2}an=6an−1−9an−2, which is the general 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)9na_n = (c_1 + c_2 n) 9^nan=(c1+c2n)9n
an=c19n+c2na_n = c_1 9^n + c_2 nan=c19n+c2n