# Task 3. (12 points)

The sequence $a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots$ is such that $a_{2 n}=\frac{1}{a_{2 n-1}}$, and $a_{2 n+1}=1-a_{2 n}$.

Find $a_{1}$, if $a_{2018}=2$.
# Solution.

Let's find several of the last values of the sequence, considering that $a_{2 n-1}=\frac{1}{a_{2 n}}$ and $a_{2 n}=1-a_{2 n+1}:$

$$
a_{2018}=2 ; a_{2017}=0.5 ; a_{2016}=0.5 ; a_{2015}=2 ; a_{2014}=-1 ; a_{2013}=-1 ; a_{2012}=2 ; a_{2011}=0.5 \ldots
$$

Thus, we have obtained a periodic sequence with a period of 6.

Therefore, $a_{1}=a_{1+3 \times 6.6}=a_{2017}=0.5$.

## Answer. 0.5

| Criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete solution. | + | 12 |
| The main idea of the solution is found. There is a lack of necessary rigor in obtaining the answer. The answer is correct. | $\pm$ | 9 |
| The main idea of the solution is found. There is a lack of necessary rigor in obtaining the answer and/or a computational error is made. The answer may be incorrect. | $+/ 2$ | 6 |
| The idea of periodicity is shown. The period is not found or is found incorrectly. | $\mp$ | 2 |
| The solution does not meet any of the criteria described above. | $-/ 0$ | 0 |
| Maximum score |  | 12 |