If $a_n+b_n\sqrt{3}=(2+\sqrt{3})^n$, then what's $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}$? Let $a_n$ and $b_n$ be integers defined in the following way: $$a_n+b_n\sqrt{3}=(2+\sqrt{3})^n.$$ Compute $$\lim_{n\rightarrow\infty}\frac{a_n}{b_n}.$$
I tried expanding using binomial theorem:
$$ (2+\sqrt{3})^n=\binom{n}{0}2+\binom{n}{1}2^2\sqrt{3}+\binom{n}{2}2^3\cdot 3+\ldots+\binom{n}{n-1}2\sqrt{3}^{n-1}+\binom{n}{n}\sqrt{3}^{n}$$
and I think we have that: $$ a_n=\binom{n}{0}2+\binom{n}{2}2^3\cdot 3+\ldots+\binom{n}{n}\sqrt{3}^{n}$$
$$ b_n=\binom{n}{1}2^2\sqrt{3}+\ldots+\binom{n}{n-1}2\sqrt{3}^{n-1}$$
But, frankly, I do not know what to do next. 

For fun, you can also do this with linear algebra. Let $V = \mathbb{Q}(\sqrt{3})$ as a $\mathbb{Q}$-vector space with basis $\{1,\sqrt{3}\}$. The "multiplication-by-$(2+\sqrt{3})$" map looks like $\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix},$ so $$\begin{pmatrix} a_n \\ b_n \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}.$$ In the limit, the largest eigenvalue of $\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$ dominates, so $\begin{pmatrix} a_n \\ b_n \end{pmatrix}$ will be close to a multiple of the eigenvector $\begin{pmatrix} \sqrt{3} \\ 1 \end{pmatrix}$; i.e. $\frac{a_n}{b_n} \rightarrow \sqrt{3}.$
