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

Question

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

Answer 1

Induction.

True for $n=0,n=1$. We will assume true up to $n$, will show holds for $n+1$: $$a_{n+1} = 6 a_n -9 a_{n-1} = 6 n 3^{n-1} - 9 (n-1) 3^{n-2}=(2n -(n-1))3^n=(n+1)3^{(n+1)-1}.$$

Answer 2

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 $a_{n+3}$ =$3^{n+2}(n+3)$