Given $ a_n= 6a_{n-1} -4a_{n-2}$ and initial values, find a closed form for $a_n$ I have a recursive formula:
$$ a_n= 6a_{n-1} -4a_{n-2}$$
with $a_0=1$ and $a_1=3$, and I need to find a closed-form expression of $(a_n)_{n\in \mathbb{N}}$.
I managed to calculate almost everything but at the end I get this expression:
$$ a_n= \frac{(3+\sqrt{5})^n}{2} + \frac{(3-\sqrt{5})^n}{2} $$
Is there a way to prove the following statement? Because Everything I have tried up till now doesn't do the job, and are these two expressions equal at all?
$$ \frac{(3+\sqrt{5})^n}{2} + \frac{(3-\sqrt{5})^n}{2} = \left \lceil \frac{(3+\sqrt5)^n}{2} \right \rceil$$

Note that $3-\sqrt5$ is a little less than $0.764$, so $0<\frac{(3-\sqrt5)^n}2<\frac12$ for all $n\ge 1$. The lefthand side of your final expression must be an integer; call it $m$. Thus,
$$0<m-\frac{(3+\sqrt5)^n}2<\frac12\;,$$
and it follows immediately that
$$m=\left\lceil\frac{(3+\sqrt5)^n}2\right\rceil\;.$$
For this it’s actually enough that $\frac{(3-\sqrt5)^n}2<1$; the fact that it’s less than $\frac12$ allows the stronger conclusion that $a_n$ is actually the integer closest to $\frac{(3+\sqrt5)^n}2$ for $n\ge 1$.
