Solving $a_n=3a_{n-1}-2a_{n-2}+3$ for $a_0=a_1=1$ I'm trying to solve the recurrence $a_n=3a_{n-1}-2a_{n-2}+3$ for $a_0=a_1=1$.  First I solved for the homogeneous equation $a_n=3a_{n-1}-2a_{n-2}$ and got $\alpha 1^n+\beta 2^n=a_n^h$.  Solving this gives $a_n^h =1$.  The particular solution, as I understand, will be $a_n^*=B$ since $f(n)=3\times 1^n$.  But then I get $B=a^*=3a^*_{n-1}-2a^*_{n-2}+3=B+3$.  This has to be a mistake, but I don't see what I did wrong.

$$a_n - a_{n-1}+3=2(a_{n-1}-a_{n-2}+3)$$
$$a_n - a_{n-1}+3=2^{n-1}(a_1-a_0+3)=3\cdot 2^{n-1}$$
$$\\$$
$$a_n = a_{n-1} + 3\cdot (2^{n-1}-1)$$
$$a_{n-1} = a_{n-2} + 3\cdot (2^{n-2}-1)$$
$$...$$
$$a_1 = a_0 + 3\cdot (2^0-1)$$
Adding side by side,
$$a_n=a_0+3\cdot (2^0+2^1+...+2^{n-1}-n)$$
$$=1+3\cdot (2^n-1-n)=3\cdot2^n-3n-2$$
