\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{n.\left(n+3\right)}\)=\(\frac{1}{3}\)(\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{n}+\frac{1}{n+3}\))=\(\frac{1}{5}-\frac{1}{n+3}=\frac{101}{1540}:\frac{1}{3}=\frac{303}{1540}\)
=>\(\frac{1}{n+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)vậy n= 308+3=311