Given the following recursive function: $$a_{n+1}=a_{n}+2b_{n}$$ $$b_{n+1}=a_{n}+b_{n}$$ With $a_1 = 3$ and $b_1 = 2$, I'm trying to figure out how to simplify the function in order to calculate $a_n$ for a large value of $n$ and haven't succeeded.
If there is a method to simplify $a_n$ or if you can induce a pattern of any sort please post it. Thanks for your help.
From $a_{n+1}=a_n+2b_n$, we get $$b_n=\frac12a_{n+1}-\frac12a_n$$and by incrementing $n$ by $1$, we also get $$\begin{align}a_{n+2}&=a_{n+1}+2b_{n+1}\\&=a_{n+1}+2a_n+2b_n\\&=a_{n+1}+2a_n+a_{n+1}-a_n\\&=2a_{n+1}+a_n\end{align}$$ So we get a homogeneous second-order recurrence relation. $$a_{n+2}-2a_{n+1}-a_n=0$$ Since the solutions to $$\lambda^2-2\lambda-1=0$$ are $\lambda=1\pm2$, we get that the general solution of the recurrence relation is $$a_n=A\left(1+\sqrt2\right)^n+B\left(1-\sqrt2\right)^n$$ Since $a_1=3$ and $a_2=7$, we get that $$A=\frac{1+\sqrt2}{2}\\B=\frac{1-\sqrt2}{2}$$ Therefore, $$\begin{align}a_n&=\frac{1+\sqrt2}{2}\left(1+\sqrt2\right)^n+\frac{1-\sqrt2}{2}\left(1-\sqrt2\right)^n\\&=\boxed{\frac12\left({1+\sqrt2}\right)^{n+1}+\frac12\left({1-\sqrt2}\right)^{n+1}}\end{align}$$ As you can see, as $n\rightarrow\infty$, the $\frac12\left({1-\sqrt2}\right)^{n+1}$ term tends to $0$, so $$a_n\approx\frac12\left({1+\sqrt2}\right)^{n+1}$$