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TR-MO-R1 2024 Problem 31

    Bir {an}n≥1\{a_n\}_{n\geq 1} dizisi, a1=2a_1 = 2 ve her n≥1n \geq 1 tam sayısı için

    (an+1−3an+1an+3)(an+1−5an+1an+5)=0\left(a_{n+1} - \frac{3a_n + 1}{a_n + 3}\right)\left(a_{n+1} - \frac{5a_n + 1}{a_n + 5}\right) = 0

    şartını sağlamaktadır. Buna göre, a2024a_{2024} sayısı kaç farklı değer alabilir?

    a) 20242024 b) 202422024^2 c) 220232^{2023} d) 220242^{2024} e) Hiçbiri

    Dataset text TÜBİTAK olympiad set, Matematik_2024_1.Asama_31. Quoted with source.

    A model read this text from the scan of the paper by OCR, with vision/claude@2.1.318 (claude-opus-5-5). No person has checked it against the paper yet.

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