The question is:
Determine the numbers $a_n$ for $n \geq 0$ that satisfy the recurrence relation $a_{n+2} = 4a_{n+1} - 2a_n$ for $n\geq 0$. with Boundary conditions $a_0 = 0$ and $a_1 = 1$.
Now my base step would be to fill in the formula : $$ \begin{align} a_0 :a_2&=4a_1-2a_0 \\ &=4*1 -2*0 =4 \end{align} $$ So we assume it holds for $n = 0$, and thus for $k$. fill in $k+1$: $$ \begin{align} a_{k+3}&=4a_{k+2}-2a_{k+1}\\ \end{align} $$
Which doesnt really help me at all. How do I tackle this? AM I looking at this the wrong way?
Since the roots are of the form $2\pm\sqrt{2}$, $a_n=A(2+\sqrt{2})^n+B(2-\sqrt{2})^n$
To see why this works, notice that
$ \left( \begin{array} [cc] a_{n+2} && a_{n+1} \\ a_{n+1} && a_{n} \end{array} \right)=\left( \begin{array} [cc] \\ 4 && -2 \\ 1 && 0 \end{array} \right)\left( \begin{array} [cc] a_{n+1} && a_{n} \\ a_{n} && a_{n-1} \end{array} \right)$ and diagonalise $\left( \begin{array} [cc] \\ 4 && -2 \\ 1 && 0 \end{array} \right)$.