Lời giải:
$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n(n+1)}=\frac{2022}{2023}$
$\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{n(n+1)}=\frac{2022}{2023}$
$2[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{n(n+1)}]=\frac{2022}{2023}$
$2[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{n(n+1)}]=\frac{2022}{2023}$
$2(\frac{1}{2}-\frac{1}{n+1})=\frac{2022}{2023}$
$1-\frac{2}{n+1}=1-\frac{1}{2023}$
$\Rightarrow \frac{2}{n+1}=\frac{1}{2023}$
$\Rightarrow n+1=2.2023=4046$
$\Rightarrow n=4045$