Olympiad Maths Prep

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Problem 789

AIME late
Algebra Difficulty 5.4 Find the answer

Task 4. (20 points) For the numerical sequence {xn}\left\{x_{n}\right\}, all terms of which, starting from n≥2n \geq 2, are distinct, the relation xn=xn−1+298xn+xn+1300x_{n}=\frac{x_{n-1}+298 x_{n}+x_{n+1}}{300} holds. Find x2023−x22021⋅2022x2023−x1−2023\sqrt{\frac{x_{2023}-x_{2}}{2021} \cdot \frac{2022}{x_{2023}-x_{1}}}-2023.

Official solution

# Solution.

From the given relations in the problem, it is easily deduced that for all n≥2n \geq 2, xn−xn−1=xn+1−xnx_{n}-x_{n-1}=x_{n+1}-x_{n}, which implies that the sequence is an arithmetic progression. Indeed,

xn=xn−1+298xn+xn+13002xn=xn−1+xn+1xn−xn−1=xn+1−xn \begin{gathered} x_{n}=\frac{x_{n-1}+298 x_{n}+x_{n+1}}{300} \\ 2 x_{n}=x_{n-1}+x_{n+1} \\ x_{n}-x_{n-1}=x_{n+1}-x_{n} \end{gathered}

Let the common difference of this progression be d,d≠0d, d \neq 0 (as per the condition).

Then x2023−x22021⋅2022x2023−x1−2023=x1+2022d−x1−d2021⋅2022x1+2022d−x1−2023=\sqrt{\frac{x_{2023}-x_{2}}{2021} \cdot \frac{2022}{x_{2023}-x_{1}}}-2023=\sqrt{\frac{x_{1}+2022 d-x_{1}-d}{2021} \cdot \frac{2022}{x_{1}+2022 d-x_{1}}}-2023= =2021d2021⋅20222022d−2023=1−2023=−2022=\sqrt{\frac{2021 d}{2021} \cdot \frac{2022}{2022 d}}-2023=1-2023=-2022.

Answer. -2022.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.