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HMMT 2016 A5

  • Algebra
  • Answer

An infinite sequence of real numbers a1,a2,…a_{1}, a_{2}, \ldots satisfies the recurrence

an+3=an+2−2an+1+ana_{n+3}=a_{n+2}-2 a_{n+1}+a_{n}

for every positive integer nn. Given that a1=a3=1a_{1}=a_{3}=1 and a98=a99a_{98}=a_{99}, compute a1+a2+⋯+a100a_{1}+a_{2}+\cdots+a_{100}. Proposed by: Evan Chen

Dataset text NuminaMath olympiads. Quoted with source.

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