<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] $$x_{n+1} = 3x_{n} + \sqrt{8x_{n}^2+2}$$
$$x_{n+1}-3x_{n} = \sqrt{8x_{n}^2+2}$$
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 -
$$x_{2020}=5$$
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