HMMT February, 2016

Algebra Difficulty 5.2 AIME, harder Prove it United States

Problem:

An infinite sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the recurrence
$a_{n+3}=a_{n+2}-2 a_{n+1}+a_{n}$
for every positive integer $n$ . Given that $a_{1}=a_{3}=1$ and $a_{98}=a_{99}$ , compute $a_{1}+a_{2}+\cdots+a_{100}$ .

Proposed by: Evan Chen

Solution

Solution:

$\begin{aligned} \sum_{k=1}^{n} a_{k} & =a_{1}+a_{2}+a_{3}+\sum_{k=1}^{n-3}\left(a_{k}-2 a_{k+1}+a_{k+2}\right) \\ & =a_{1}+a_{2}+a_{3}+\sum_{k=1}^{n-3} a_{k}-2 \sum_{k=2}^{n-2} a_{k}+\sum_{k=3}^{n-1} a_{k} \\ & =2 a_{1}+a_{3}-a_{n-2}+a_{n-1} \end{aligned}$

Putting $n=100$ gives the answer.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.