Question

AJUTORRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR! ​

Answer 1

$\displaystyle\it\\a=\frac{1}{3}\bigg(\frac{7}{4}+\frac{8}{5}+\frac{9}{6}+...+\frac{2020}{2017}\bigg).\\b=\frac{1}{3}\bigg(\frac{1}{4}+\frac{2}{5}+\frac{3}{6}+...+\frac{2014}{2017}\bigg).\\------------(+)--\\a+b=\frac{1}{3}\bigg(\frac{8}{4}+\frac{10}{5}+\frac{12}{6}+...+\frac{4034}{2017}\bigg)=\frac{1}{3}\bigg(\underbrace{\it 2+2+2+...+2}_{2~de~2014~ori}\bigg)=\boxed{\it\frac{2014}{3}}.$

$\displaystyle\it\\\frac{3}{2014}\big(a+b\big)=\frac{3}{2014}\cdot\frac{2014}{3}=\boxed{\it 1\in\mathbb{N}}.$