Prove $a_n = n 3^{n-1}$, recurrence relation. 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

First off let us see if the base case holds.
$a_{0}$=$0*3^n-1$=0 that means $a_{0}$ holds
$a_{1}$=$1*3^0=1$ that means $a_{1}$ holds too
Then to do the inductive step we have to see if n+1 holds.
$a_{n+3}$=$6(n+2)*3^n$ + $2$ - $1$ - $9(n+1)*3^n$
=$2(n+2)*3^n+2$-$(n+1)*3^n+2$=$3^n+2(2(n+2)-n+1)$= $3^{n+2}(n+3)$
That means that the inductive step holds, since if we subsitiute n+3 into our formula we get $a_{n+3}$ =$3^{n+2}(n+3)$
