Answer to Question #150275 in Algebra for Khan Asf

Question

The sequence is given by the formula X(n+1) = 4Xn + √((15Xn^2)+(2)). It is known that X2017 + X2023 = 1464. Find X2020.

Answer 1

It is known that $X_{n+1}=4X_n+\sqrt{15X_n^2+2}$ and $X_{2017}+X_{2023}=1464$ .

Note that $X_{n+1}= 4X_n+\sqrt{15X_n^2+2}>4X_n+\sqrt{9X_n^2}=4X_n+3|X_n|\geq 4X_n-3X_n=X_n$

It means that this sequence is strictly increasing and $X_n\cancel{=}X_k$ for all $n\cancel{=} k$ .

Using the formula of the $n$ th term, we get:

$X_{n+1}-4X_n=\sqrt{15X_n^2+2}$

$X_{n+1}^2-8X_nX_{n+1}+16X_n^2=15X_n^2+2$

$X_{n+1}^2-8X_nX_{n+1}+X_n^2=2$

If we consider $n=k$ and $n=k-1$ , we get:

$\begin{cases} X_{k}^2-8X_{k-1}X_{k}+X_{k-1}^2=2 \\ X_{k+1}^2-8X_kX_{k+1}+X_k^2=2 \\ \end{cases} \quad \Rightarrow X_{k+1}^2-X^2_{k-1}-8X_{k}(X_{k+1}-X_{k-1})=0$

or $(X_{k+1}-X_{k-1})(X_{k+1}+X_{k-1}-8X_k)=0$ .

Therefore, $X_{k+1}+X_{k-1}-8X_k=0$ for all $k$ . (1)

Let us consider $k=2020:$

$X_{2021}+X_{2019}=8X_{2020}$ .

Using formula (1) for $k=2021$ and $k=2019$ , we get:

$\frac{X_{2020}+X_{2022}}{8}+\frac{X_{2018}+X_{2020}}{8}=8X_{2020}$

$X_{2022}+X_{2018}=62X_{2020}$

Using formula (1) for $k=2022$ and $k=2018$, we get:

$\frac{X_{2021}+X_{2023}}{8}+\frac{X_{2017}+X_{2019}}{8}=62X_{2020}$

$(X_{2021}+X_{2019})+(X_{2023}+X_{2017})=496X_{2020}$

$8X_{2020}+1464=496X_{2020}$

$488X_{2020}=1464$

$X_{2020}=3$

Answer: $X_{2020}=3.$