<s>[INST] 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.
 [/INST] 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 $x_n = \frac{990}{6(6^2-3)} = 5$.
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