Consider the recurrence an=4an−1−4an−2a_n = 4a_{n-1} - 4a_{n-2}an=4an−1−4an−2 with a0=1,a1=2a_0 = 1, a_1 = 2a0=1,a1=2. What is the general solution?
an=2na_n = 2^nan=2n
an=(A+Bn)2na_n = (A+Bn)2^nan=(A+Bn)2n
an=A(2n)+B(2n)a_n = A(2^n) + B(2^n)an=A(2n)+B(2n)
an=n⋅2na_n = n \cdot 2^nan=n⋅2n