Let $k \in \mathbb{N}$ with $k \geq 2$.
$x_k=\frac{2}{3}x_{k-1}+\frac{1}{3}x_{k-2}$
$x_0=0$
$x_1=1$
How to find the solution for $\lbrace x_k \rbrace _{k \in \mathbb{N_0}}$?
Since it's for $k \geq 2$ I got:
$k=2:$
$x_2=\frac{2}{3}x_1+\frac{1}{3}x_0=\frac{2}{3}$
But how to continue to find the solution? I thought about using induction and I get as result:
$x_{k+1}= \frac{2}{3}x_k+\frac{1}{3}x_{k-1}$
Here I don't know what to do next.
Write $$ x_k=\frac{2}{3}x_{k-1}+\frac{1}{3}y_{k-1}\\ y_k = x_{k-1} $$ so $$ \left(x_k \atop y_k \right)=\left({\frac23 \quad \frac13 } \atop {1 \quad 0 } \right) \left(x_{k-1} \atop y_{k-1} \right) $$ which gives you $$ \left(x_k \atop y_k \right)=\left({\frac23 \quad \frac13 } \atop {1 \quad 0 } \right)^k \left(x_{0} \atop x_1 \right) $$ So you need the powers of the matrix which can be found by eigenvalues. They are $1$ and $ - \frac13$. So you can as well make the ansatz
$$ x_k = a + b (-\frac13)^k $$ solving for $a$ and $b$ gives $$ 0 = x_0 = a + b \\ 1 = x_1 = a + -\frac{b}{3} $$ So $$ x_k = \frac34 (1 - (-\frac13)^k) $$