## Solution.

We have $\quad a_{2}=2 a_{1}^{2}-a_{0}=2\left(\frac{1+\sqrt{3}}{2 \sqrt{2}}\right)^{2}-1=\frac{\sqrt{3}}{2}, \quad a_{3}=2 a_{1} a_{2}-a_{1}=2 \cdot \frac{\sqrt{3}}{2} \cdot\left(\frac{1+\sqrt{3}}{2 \sqrt{2}}\right)-\frac{1+\sqrt{3}}{2 \sqrt{2}}=\frac{\sqrt{2}}{2}$, $a_{4}=2 a_{1} a_{3}-a_{2}=2 \cdot \frac{\sqrt{2}}{2} \cdot\left(\frac{1+\sqrt{3}}{2 \sqrt{2}}\right)-\frac{\sqrt{3}}{2}=\frac{1}{2}$. These values reasonably suggest the hypothesis $a_{n}=\cos \left(\frac{\pi}{12} n\right), \forall n \geq 2$. We will prove its correctness using mathematical induction. Indeed, we have

$$
\cos \left(\frac{\pi}{12} \cdot 2\right)=\cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}=a_{2}
$$

Assume the statement is true for $\forall n<N, N \geq 2$. Let's prove the correctness of the statement for $n=N$. Indeed, we have

$$
\begin{aligned}
& a_{N}=2 a_{1} a_{N-1}-a_{N-2}=2 \cos \frac{\pi}{12} \cdot \cos \left(\frac{\pi}{12}(N-1)\right)-\cos \left(\frac{\pi}{12}(N-2)\right)= \\
& =\cos \left(\frac{\pi}{12}(N-1)+\frac{\pi}{12}\right)+\cos \left(\frac{\pi}{12}(N-1)-\frac{\pi}{12}\right)-\cos \left(\frac{\pi}{12}(N-2)\right)= \\
& =\cos \left(\frac{\pi}{12} N\right)+\cos \left(\frac{\pi}{12}(N-2)\right)-\cos \left(\frac{\pi}{12}(N-2)\right)=\cos \left(\frac{\pi}{12} N\right)
\end{aligned}
$$

By the method of mathematical induction, we have $a_{n}=\cos \left(\frac{\pi}{12} n\right), \forall n \geq 2$.

Since the cosine function is periodic with period $2 \pi$, it follows that the sequence $\left(a_{n}\right)_{n=0}^{\infty}$ is periodic with period 24. Therefore, $a_{2020}=a_{2020 \bmod 24}=a_{4}=\frac{1}{2}$.

We calculate $\lim _{n \rightarrow \infty} \frac{a_{n}}{n^{2}}$. We obtain $0 \leq\left|\frac{a_{n}}{n^{2}}\right|=\left|\frac{1}{n^{2}} \cos \left(\frac{\pi}{12} n\right)\right| \leq \frac{1}{n^{2}}$. Taking the limit as $n \rightarrow \infty$, we get $0 \leq \lim _{n \rightarrow \infty}\left|\frac{a_{n}}{n^{2}}\right| \leq \lim _{n \rightarrow \infty} \frac{1}{n^{2}}=0$, i.e., $\lim _{n \rightarrow \infty}\left|\frac{a_{n}}{n^{2}}\right|=0$, which implies $\lim _{n \rightarrow \infty} \frac{a_{n}}{n^{2}}=0$.
11.6. Let the sequence $\left(a_{n}\right)_{n=0}^{\infty}$ be defined by the relations $a_{0}=1, a_{1}=\frac{1+\sqrt{3}}{2 \sqrt{2}}$ and $a_{n}=2 a_{1} a_{n-1}-a_{n-2}, \forall n \geq 2$. Calculate the value of $a_{2020}$ and determine $\lim _{n \rightarrow \infty} \frac{a_{n}}{n^{2}}$.