Consider the recurrence an=4an−1−4an−2a_n = 4a_{n-1} - 4a_{n-2}an=4an−1−4an−2. What is the general solution?
an=c12n+c22na_n = c_1 2^n + c_2 2^nan=c12n+c22n
an=(c1+c2n)2na_n = (c_1 + c_2 n) 2^nan=(c1+c2n)2n
an=c14n+c21na_n = c_1 4^n + c_2 1^nan=c14n+c21n
an=c1n2n+c2n22na_n = c_1 n 2^n + c_2 n^2 2^nan=c1n2n+c2n22n