Solve for the general term of an=2an−1+8an−2a_n = 2a_{n-1} + 8a_{n-2}an=2an−1+8an−2 with a0=0,a1=4a_0 = 0, a_1 = 4a0=0,a1=4.
an=423((1+9)n−(1−9)n)a_n = \frac{4}{2\sqrt{3}} ( (1+\sqrt{9})^n - (1-\sqrt{9})^n )an=234((1+9)n−(1−9)n)
an=13(4n−(−2)n)a_n = \frac{1}{\sqrt{3}} ( 4^n - (-2)^n )an=31(4n−(−2)n)
an=23(4n−(−2)n)a_n = \frac{2}{\sqrt{3}} ( 4^n - (-2)^n )an=32(4n−(−2)n)
an=2n−(−4)na_n = 2^n - (-4)^nan=2n−(−4)n