[tex]\displaystyle\\
\text{Daca }~n=2017, \text{ atunci }~\frac{n+1}{n}= \frac{2017+1}{2017}= \boxed{\frac{2018}{2017}} \\\\\\
A=\left\{ \frac{1}{2};\frac{2}{3};\frac{3}{4};\cdots;\frac{2015}{2016};\frac{2016}{2017}; \frac{2017}{2018} \right\}\\\\
B=\left\{\frac{2}{1};\frac{3}{2};\frac{4}{3};\cdots;\frac{2016}{2015};\frac{2017}{2016}; \frac{2018}{2017} \right\}
[/tex]
[tex]\displaystyle\\
A\cup B=\left\{ \frac{1}{2};\frac{2}{3};\frac{3}{4};...;\frac{2015}{2016};\frac{2016}{2017}; \frac{2017}{2018};\frac{2}{1};\frac{3}{2};\frac{4}{3};...;\frac{2016}{2015};\frac{2017}{2016}; \frac{2018}{2017} \right\}\\\\
\text{Observam ca fiecare multime are cate 2017 termeni. }\\
\text{Multimea } A\cup B \text{ are 2017 + 2017 = 4034 termeni.}\\
\text{Calculam suma elementelor multimii } A\cup B.\\
\text{Pentru inceput adunam intre ei termenii care au acelasi numitor.}
[/tex]
[tex]\displaystyle\\
S = \frac{2}{1}+\left(\frac{1}{2}+\frac{3}{2} \right)+\left( \frac{2}{3}+\frac{4}{3}\right)+\left(\frac{3}{4}+\frac{5}{4} \right)+....+\left( \frac{2014}{2015}+\frac{2016}{2015}\right)+\\\\
+\left( \frac{2015}{2016}+\frac{2017}{2016} \right)+\left(\frac{2016}{2017}+\frac{2018}{2017} \right) +\frac{2017}{2018}\\\\
\Rightarrow \text{\bf 2016 sume + 2 termeni}\\\\
S=\frac{2}{1}+\frac{4}{2}+ \frac{6}{3}+\frac{8}{4}+...+ \frac{4030}{2015}+ \frac{4032}{2016}+\frac{4034}{2017}+\frac{2017}{2018}
[/tex]
[tex]\displaystyle\\S = \underbrace{2+2+2+2+...+2+2+2}_{\texttt{\bf 2017 de 2}} + \frac{2017}{2018} \\\\
S = 2017 \times 2 + \frac{2017}{2018} = 4034 +\frac{2017}{2018} = \boxed{4034\texttt{ \bf intregi si }\frac{2017}{2018}}[/tex]
