A sequence $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and find $x_{2020}$ The sequence is given by the formula $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and it is known that  $x_{2017} + x_{2023} = 990$, then what is $x_{2020}$ = ?
My little approch:
It is given that  $x_{2017} + x_{2023} = 990$ ----- (1)  and $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$
So,
$x_{2017} = x_{2016+1} = 3 x_{2016} + \sqrt{8x^2_{2016} + 2}$ -------(2)
$x_{2023} = x_{2022+1} = 3 x_{2022} + \sqrt{8x^2_{2022} + 2}$ -------(3)
Then from (1),(2),(3) =>
$3 x_{2016} + \sqrt{8x^2_{2016} + 2} + 3 x_{2022} + \sqrt{8x^2_{2022} + 2} = 990$
$3 (x_{2016}+x_{2022} ) + \sqrt{8x^2_{2016} + 2}  + \sqrt{8x^2_{2022} + 2} = 990$
$3 (x_{2016}+x_{2020} + x_{2021}) + \sqrt{8x^2_{2016} + 2}  + \sqrt{8x^2_{2020} + 2}+ \sqrt{8x^2_{2021} + 2}+ \sqrt{8x^2_{2022} + 2} = 990$
I stuck here and can't go further. Please help me with it.
 A: We have
$$x_{n+1}-3x_n = \sqrt{8x_n^2+2}$$
$$(x_{n+1}-3x_n)^2 = 8x_n^2+2$$
$$x_{n+1}^2 -6x_{n+1} x_n + x_n^2 = 2   \tag{I}$$
And because
$$x_{n-1}^2 -6x_{n-1} x_n + x_n^2 = 2   \tag{II}$$
From $(I)$ and $(II)$, we have, for all $n$
$$(x_{n+1}-x_{n-1})(x_{n+1}+x_{n-1}-6x_{n})=0$$
Is it possible that $x_{n+1}=x_{n-1}$ ? No, because $x_n$ is an increasing series. Effectively, we have $x_{n+1}-x_n = \sqrt{8x_n^2+2}+2x_n > \sqrt{4x_n^2}+2x_n \ge 0$.
So, $x_{n+1}+x_{n-1}-6x_{n}=0$ for all $n$ or $x_n = \frac{x_{n+1}+x_{n-1}}{6}$
Now, we
$$ 
\begin{align}
x_n &= \frac{x_{n-1}+x_{n+1}}{6} \\
&= \frac{ \frac{x_{n-2}+x_{n}}{6} +\frac{x_{n}+x_{n+2}}{6}}{6} \\
&= \frac{x_{n-2}+x_{n+2}}{6^2} + \frac{2x_n}{6^2}  \\
&= \frac{\frac{x_{n-3}+x_{n-1}}{6} +\frac{x_{n+1}+x_{n+1}}{6}}{6^2} + \frac{2x_n}{6^2}  \\
&= \frac{x_{n-3}+x_{n+3}}{6^3}+\frac{x_{n-1}+x_{n+1}}{6^3} + \frac{2x_n}{6^2}  \\
&= \frac{x_{n-3}+x_{n+3}}{6^3}+\frac{x_n}{6^2} + \frac{2x_n}{6^2}  \\
\end{align}
$$
So,
$$(1-\frac{3}{6^2})x_n= \frac{x_{n-3}+x_{n+3}}{6^3}$$
Or
$$x_n = \frac{x_{n-3}+x_{n+3}}{6(6^2-3)}$$
Hence, if $x_{2017}+x_{2023} = 990$, then $x_n = \frac{990}{6(6^2-3)} = 5$.
A: $$x_{n+1} = 3x_{n} + \sqrt{8x_{n}^2+2}$$
$$x_{n+1}-3x_{n} = \sqrt{8x_{n}^2+2}$$
By squaring both sides,
$$x_{n+1}^2+x_{n}^2-6x_{n}x_{n+1}=2$$
Since this is a general expression, we can write -
$$x_{n+1}^2+x_{n}^2-6x_{n}x_{n+1}=x_{n}^2+x_{n-1}^2-6x_{n-1}x_{n}$$
Simplifying this we get,
$$x_{n+1}+x_{n-1}=6x_{n}$$
Now our task is to find $x_n$ for $n=2020$ when $x_{n+3}+x_{n-3}$ is given.
First we will find $x_{n+2}+x_{n-2}$.
Let $$x_{n+2}+x_{n-2}=y$$.
By adding $2x_{n}$ to both sides of the equation,
$$6(x_{n+1}+x_{n-1})=y+2x_{n}$$
So,
$$y=34x_{n}$$
Now let,
$$x_{n+3}+x_{n-3}=z$$
Add $x_{n+1}+x_{n-1}$ to both the sides,
$$6(x_{n+2}+x_{n-2})=z+6x_{n}$$
$$z=198x_{n}$$
So, we get -
$$x_{2020}=5$$
A: After you get $x_{n+1} - 6x_n + x_{n-1}=0$ as in the other two answers, a conceptually  easy way to establish the relationship between $x_n$ and $x_{n-3}+x_{n+3}$ is as follows:
Rewrite the recurrence equation as
$$(\mathbb E^2-6\mathbb E+1)x_n=0$$
where $\mathbb E$ is the forward shift operator $\mathbb E^{i} x_n = x_{n+i}, \forall i \in \mathbb N$. Then the characteristic equation is $$\lambda^2 - 6\lambda + 1 =0$$ with two roots $a, b$ such that $a+b=6, ab=1$. Now
$$a^3\cdot b^3 =1, a^3+b^3 = (a+b)((a+b)^2-3ab)=6\cdot (6^2-3)=198$$
Therefore $$\lambda^6 - 198 \lambda^3+1=(\lambda^3-a^3)(\lambda^3-b^3)=g(\lambda)(\lambda-a)(\lambda-b)=g(\lambda)(\lambda^2-6\lambda+1)$$
where $g(\lambda)=(\lambda^2 + a\lambda +a^2)(\lambda^2 + b\lambda +b^2)$ (we don't need to know the coefficients, we only need the fact that $g$ is a polynomial)
Then
$$(\mathbb E^6 - 198 \mathbb E^3 + 1) x_n = g(\mathbb E)(\mathbb E^2 - 6\mathbb E+1)x_n = 0\\
\implies x_{n+6} - 198 x_{n+3} + x_n=0 \implies x_{2020} = 990/198=5.$$
