A recurrence relation an=4an−1−4an−2a_n = 4a_{n-1} - 4a_{n-2}an=4an−1−4an−2 has the general solution (c1+c2n)2n(c_1 + c_2 n) 2^n(c1+c2n)2n. If a0=1,a1=4a_0 = 1, a_1 = 4a0=1,a1=4, what is c1c_1c1 and c2c_2c2?
c1=1,c2=1c_1 = 1, c_2 = 1c1=1,c2=1
c1=1,c2=0c_1 = 1, c_2 = 0c1=1,c2=0
c1=0,c2=1c_1 = 0, c_2 = 1c1=0,c2=1
c1=2,c2=1c_1 = 2, c_2 = 1c1=2,c2=1