Solve the recurrence an=2an−1+2n−1a_n = 2a_{n-1} + 2^{n-1}an=2an−1+2n−1 with a0=1a_0 = 1a0=1.
an=(n+1)2n−1a_n = (n+1)2^{n-1}an=(n+1)2n−1
an=2n+n2n−1a_n = 2^n + n2^{n-1}an=2n+n2n−1
an=(n+2)2n−2a_n = (n+2)2^{n-2}an=(n+2)2n−2
an=2n+2n−1a_n = 2^n + 2^{n-1}an=2n+2n−1