A sequence $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and find $x_{2020}$

Question

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.

Answer 1

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 $x_n = \frac{990}{6(6^2-3)} = 5$.

Answer 2

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

Answer 3

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 $$(\mathbb E^6 - 198 \mathbb E^3 + 1) x_n = g(\mathbb E)(\mathbb E^2 - 6\mathbb E+1)x_n = 0\\ \implies x_{n+6} - 198 x_{n+3} + x_n=0 \implies x_{2020} = 990/198=5.$$