# PROBLEM 1 

The following sequence of numbers is given: $a_{1}=1+1^{-1}, a_{2}=1+2^{-1}, a_{3}=1+3^{-1}, \ldots, a_{2016}=1+2016^{-1}$.

a) Compare $a_{2016} \mathrm{with} a_{2011}$.

b) Show that the product $a_{1} \cdot a_{2} \cdot a_{3} \cdot \ldots \cdot a_{2016}$ is a natural number.

c) Determine the number $n \in \mathbb{N}^{*}$, for which $a_{n} \cdot a_{n+1} \cdot a_{n+2} \in \mathbb{N}$.
## Solution:

a) $a_{2011}=1+2011^{-1}=1+\frac{1}{2011}>1+\frac{1}{2016}=1+2016^{-1}=a_{2016} \Rightarrow a_{2011}>a_{2016}$
b) $a_{1} \cdot a_{2} \cdot \ldots \cdot a_{2016}=(1+1)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right) \cdot \ldots \cdot\left(1+\frac{1}{2016}\right)=$

$=2 \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \ldots \cdot \frac{2017}{2016}=2017 \in \mathbb{N}$
c) $a_{n} \cdot a_{n+1} \cdot a_{n+2}=\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n+1}\right)\left(1+\frac{1}{n+2}\right)=\frac{n+1}{n} \cdot \frac{n+2}{n+1} \cdot \frac{n+3}{n+2}=\frac{n+3}{n}=1+\frac{3}{n} \in \mathbb{N}$

$\Rightarrow n \in\{1,3\}$

$.2 p$