The recurrence an=2an−1−an−2a_n = 2a_{n-1} - a_{n-2}an=2an−1−an−2 with a0=1,a1=3a_0 = 1, a_1 = 3a0=1,a1=3 has which closed form solution?
an=1+2na_n = 1 + 2nan=1+2n
an=2n−1a_n = 2^n - 1an=2n−1
an=3n−2a_n = 3n - 2an=3n−2
an=n2+1a_n = n^2 + 1an=n2+1