I have a problem:
Let $a_0=0, a_1=1$, and let $a_{n+2}=6a_{n+1}-9a_n$ for $n\geq 0$. Prove that $a_n=n\cdot 3^{n-1}$ for all $n\geq 0$.
And I am assuming that this can be solved via induction. Only problem is I am not sure how to go about doing this. Do I use induction to prove the first part and then again on the second part?
Sorry for terrible question, I am new to this site. Thank you.
HINT:
If $a_m=m3^{m-1}$ for $m\le n+1$
using strong induction, $a_{n+2}=6(n+1)3^n-9\cdot n3^{n-1}=\cdots=(n+2)3^{(n+2)-1}$