A sequence satisfies an=4an−1−4an−2a_n = 4a_{n-1} - 4a_{n-2}an=4an−1−4an−2. What is the general solution if a0=1a_0 = 1a0=1 and a1=3a_1 = 3a1=3?
an=2n+n2n−1a_n = 2^n + n2^{n-1}an=2n+n2n−1
an=(1+n)2na_n = (1+n)2^nan=(1+n)2n
an=2n+n2na_n = 2^n + n2^nan=2n+n2n
an=3⋅2na_n = 3 \cdot 2^nan=3⋅2n