Solve the following recurrence equation
$a_0=0, a_1=7$, and $$a_n=\frac{1}{3}a_{n-1}+\frac{4}{3}a_{n-2}, n\geq2$$ I have tried using the general method, however i am getting the same thing as $a_n$ already. I am taking this class as an independent study so I am basically teaching myself.
The characteristic equation is $x^2-\frac{1}{3}x-\frac{4}{3}=0$, which we can rewrite as $(3x-4)(x+1)=0$.
So the general solution is $a_n=A(4/3)^n +B(-1)^n$, where $A$ and $B$ are constants. Use the initial conditions to find $A$ and $B$.
Since $x^2-\frac 13x-\frac 43=0\Rightarrow x=-1,\frac 43$, we have $$a_{n+1}+a_n=\frac 43\left(a_{n}+a_{n-1}\right)=\cdots=\left(\frac 43\right)^{n}(a_1+a_0),$$ $$a_{n+1}-\frac 43a_n=-\left(a_n-\frac 43a_{n-1}\right)=\cdots=(-1)^n\left(a_1-\frac 43a_0\right).$$ Then, subtract the latter from the former to get $a_n$.
