Let $a\_0 = 0, a\_1 = 1$, and let $a\_{n+2} = 6a\_{n+1} - 9a\_n$ for $n\geq 0.$ 
Prove $a\_n = n 3^{n-1}$ for $n\geq 0.$

attempt: Suppose $a\_0 = 0, a\_1 = 1$, and let $a\_{n+2} = 6a\_{n+1} - 9a\_n$.
Then notice when $n = 0$, we have $a\_2 = 6\_{1} - 9a\_0 = 6-0 = 6$.
And $a\_2 = 2\*(3^{2-1}) = 6.$
Suppose as inductive hypothesis this is true for all $n \geq 0.$ 
Then we need to show $a\_{n+1} = (n+1) 3^{(n+1)-1}= (n+1)\*3^n$.

Then $a\_{n+1} = 6a\_{n}-9a\_{n-1}$ by the recurrence relation. 
So $a\_{n+1} = 6a\_{n}-9a\_{n-1} = 6(n 3^{n-1})- 9a\_{n-1} = 3[2(n 3^{n-1})] - 3a\_{n-1}]= $ I am stuck after this, I want to show $a\_{n+1} = (n+1)\*3^n$.

Can someone please help ? Thank you