a0 = 2 & b0 = -1
3an - 2an-1 -bn-1= 0
3bn - 2bn-1 -an-1= 0
please help me in Solving this recurrence relation for an and bn with system of equations . i tried solve this with distinct roots way but i was failed , i could just approach to this :
an + bn = an-1 + bn-1
Add the two equations together, and you get
\begin{align} & (3a_n-2a_{n-1}-b_{n-1})+(3b_n-2b_{n-1}-a_{n-1})=0 \\ \implies & (a_n+b_n)-(a_{n-1}+b_{n-1})=0 \end{align}
So if we define a new sequence $c_n :=a_n+b_n$, then we know that $c_n$ satisfies
$$c_n-c_{n-1}=0 \qquad \qquad c_0=a_0+b_0=1$$
From this, we see that $c_n = 1$ for all $n$. i.e. we have
$$a_n+b_n=c_n=1 \quad \forall n \in \Bbb N \\ \implies b_n =1-a_n \quad \forall n \in \Bbb N$$
Plugging this back in to the first equation, we have
$$3a_n-2a_{n-1}-\underbrace{b_{n-1}}_{=1-a_{n-1}}=0 \implies 3a_n-a_{n-1}=1$$
The general solution is
$$a_n = A\bigg(\frac 13\bigg)^n+\frac 12$$
for some constant $A$. Applying the initial condition $a_0=2$, we find that $A=\frac 32$. Hence, the solution is
$$a_n = \frac 12 \bigg(1+\bigg(\frac 13\bigg)^{n-1}\bigg) \qquad \qquad b_n = \frac 12 \bigg(1-\bigg(\frac 13\bigg)^{n-1}\bigg)$$