Solving $a_{n+2}-6a_{n+1}+8a_n=27n^2+18$ 
Solve the recurrence relation
  $$a_{n+2}-6a_{n+1}+8a_n=27n^2+18$$

I found the homogenous solutions which are $r=2,4$ meaning that
$$a_n=C_12^n+C_24^n$$. I'm not sure what to do after that. 

Write $a_{n+2} - 6a_{n+1} + 8a_n = (a_{n+2} - 4a_{n+1})- 2(a_{n+1} - 4a_n) = 27n^2+18$. This suggests that you put $b_n = a_n - 4a_{n-1}$, then you have: $b_{n+2}-2b_{n+1} = 27n^2 + 8$. So, you now have: $b_n = (b_n - 2b_{n-1})+ 2(b_{n-1}-2b_{n-2})+ 4(b_{n-2} - 2b_{n-3})+\cdots +2^{n-2}(b_2-2b_1) + 2^{n-1}b_1 = (27(n-2)^2+8)+2(27(n-3)^2+8)+4(27(n-4)^2+8)+\cdots+2^{n-2}((a_2-4a_1)-(a_1-4a_0))+ 2^{n-1}(a_1-4a_0)$. This sum is easy to evaluate. Repeat this trick again for the $a_n$ and you can solve it without using the characteristic equation. Its a bit long but is ...fun...
