### Instruction: Define a sequence recursively by $F_{0}=0,~F_{1}=1,$ and $F_{n}$ be the remainder when $F_{n-1}+F_{n-2}$ is divided by $3,$ for all $n\geq 2.$ Thus the sequence starts $0,1,1,2,0,2,\ldots$ What is $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?$.### Response: I recognize this sequence as a variation of the Fibonacci sequence, where each term is the sum of the previous two terms, but reduced modulo 3.
I wonder if there is a pattern or cycle in this sequence that I can use to simplify the computation.
I start writing down more terms of the sequence: $0,1,1,2,0,2,2,1,0,1,1,2,0,2,2,1,0,1,\ldots$
I notice that the sequence repeats every 8 terms: $0,1,1,2,0,2,2,1.$
This means that $F_{n}$ depends only on the value of $n$ modulo 8.
For example, $F_{9}=F_{1}=1,$ $F_{10}=F_{2}=1,$ $F_{11}=F_{3}=2,$ and so on.
This also means that the sum of any 8 consecutive terms is constant: $0+1+1+2+0+2+2+1=9.$
So, to find the sum of 8 consecutive terms starting from $F_{2017},$ I just need to find the value of $F_{2017}$ and use the cycle.
To find $F_{2017},$ I need to find the remainder when $2017$ is divided by $8.$
I can use long division or a calculator to find that $2017=8\times 252+1.$
So, $2017\equiv 1\pmod{8}.$
This means that $F_{2017}=F_{1}=1.$
Therefore, the sum of 8 consecutive terms starting from $F_{2017}$ is the same as the sum of the first 8 terms: $9.$