# Task 4. (12 points)

A numerical sequence is such that $x_{n}=\frac{n+2}{n x_{n-1}}$ for all $n \geq 2$. Find the product $x_{1} x_{2} x_{3} \ldots x_{2016} x_{2017}$, if $x_{1}=1$.

#
# Solution.

Notice that $x_{n} x_{n-1}=\frac{n+2}{n}$, therefore,

$$
x_{1} x_{2} x_{3} \ldots x_{2016} x_{2017}=x_{1}\left(x_{2} x_{3}\right)\left(x_{4} x_{5}\right) \ldots\left(x_{2016} x_{2017}\right)=1 \frac{5}{3} \frac{7}{5} \ldots \frac{2017}{2015} \frac{2019}{2017}=673
$$

Answer. 673

| Criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete | + | 12 |
| Due to a computational error or typo, an incorrect answer is obtained. | $\pm$ | 9 |


| It is shown in general that $x_{n} x_{n-1}=\frac{n+2}{n}$. The answer is incorrect or missing. | $+/ 2$ | 6 |
| :---: | :---: | :---: |
| It is shown for some $n$ that $x_{n} x_{n-1}=\frac{n+2}{n}$. The answer is incorrect or missing. | $\mp$ | 2 |
| The solution does not meet any of the criteria described above. | $-/ 0$ | 0 |
| Maximum score | 12 |  |