Solve an=4an−1−3a_n = 4a_{n-1} - 3an=4an−1−3 with a0=2a_0 = 2a0=2 using back substitution. Compute a3a_3a3.
a3=189a_3 = 189a3=189
a3=60a_3 = 60a3=60
a3=193a_3 = 193a3=193
a3=255a_3 = 255a3=255