A nha
[tex]u_{n+1}-\frac{1}{n+2}=\frac{2}{3}(u_{n}-\frac{1}{n+1})[/tex]
đặt [tex]u_{n}-\frac{1}{n+1}=v_{n}[/tex]
=>[tex]v_{n+1}=\frac{2}{3}v_{n}=>v_{n}=\left (\frac{2}{3} \right )^{n-1}.v_{1}=\frac{2^{n-2}}{3^{n-1}}=>u_{n}=\frac{2^{n-2}}{3^{n-1}}+\frac{1}{n+1}=>u_{2018}=\frac{2^{2016}}{3^{2017}}+\frac{1}{2019}[/tex]