Recurrence Relationshard
0:00.0

The recurrence an=an1+2an2a_n = a_{n-1} + 2a_{n-2} with a0=1a_0 = 1 and a1=0a_1 = 0 has the general solution an=c12n+c2(1)na_n = c_1 \cdot 2^n + c_2 \cdot (-1)^n. Find a4a_4.