15. In the sequence $20,18,2,20,-18, \ldots$ the first two terms $a_{1}$ and $a_{2}$ are 20 and 18 respectively. The third term is found by subtracting the second from the first, $a_{3}=a_{1}-a_{2}$. The fourth is the sum of the two preceding elements, $a_{4}=a_{2}+a_{3}$. Then $a_{5}=a_{3}-a_{4}$, $a_{6}=a_{4}+a_{5}$, and so on.
What is the sum of the first 2018 terms of this sequence?
Solution
038

The sequence is $20,18,2,20,-18,2,-20,-18,-2,-20,18,-2,20,18, \ldots$. This is periodic and will repeat every twelve terms. The sum of the first twelve terms is 0 . Note also that $2018=12 \times 168+2$. Therefore the first 2018 terms will consist of 168 cycles of the first twelve terms with zero sum, followed by $a_{2017}=20$ and $a_{2018}=18$. Therefore the sum of the first 2018 terms is $168 \times 0+20+18=38$.