The recurrence an=an−1+2an−2a_n = a_{n-1} + 2a_{n-2}an=an−1+2an−2 with a0=1a_0 = 1a0=1 and a1=0a_1 = 0a1=0 has the general solution an=c1⋅2n+c2⋅(−1)na_n = c_1 \cdot 2^n + c_2 \cdot (-1)^nan=c1⋅2n+c2⋅(−1)n. Find a4a_4a4.
4
5
6
8