1. The sequence is defined recursively: $a_{1}=1, a_{2}=2, a_{n+2}=\frac{a_{n+1}}{a_{n}}$. Find the sum of the first 1730 terms of this sequence.
Solution. Let's find the first few terms of the sequence: $a_{3}=a_{2} / a_{1}=$ $2 / 1=2 ; a_{4}=a_{3} / a_{2}=2 / 2=1 ; a_{5}=a_{4} / a_{3}=0.5 ; a_{6}=a_{5} / a_{4}=0.5 / 1=0.5 ;$ $a_{7}=a_{6} / a_{5}=0.5 / 0.5=1$ and $a_{8}=a_{7} / a_{6}=1 / 0.5=2$. Notice that the sequence turned out to be periodic, its first six terms $1 ; 2 ; 2 ; 1 ; 0.5 ; 0.5$ will repeat infinitely. Since the sequence is periodic, with a period of six, and $1728: 6=288$ and $a_{1}+a_{2}+\ldots+a_{6}=7$, then $\sum_{i=1}^{1730} a_{i}=\sum_{i=1}^{1728} a_{i}+a_{1729}+a_{1730}=288 \cdot 7+1+2=2019$. Answer: 2019.