\(\frac{1}{21}+\frac{1}{27}+\frac{1}{36}+...+\frac{2}{n\left(n+1\right)}=\frac{2}{9}\)
\(\frac{2}{42}+\frac{2}{54}+...+\frac{2}{n\left(n+1\right)}=\frac{2}{9}\)
\(\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{n\left(n+1\right)}=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{n+1}=\frac{n+1-6}{6n+6}=\frac{1}{9}\)
\(\frac{n-5}{6n+6}=\frac{1}{9}\)
\(9n-45=6n+6\)
\(9n-6n=6+45=51\)
\(n=51:3=17\)
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{n}.\left(n+1\right)=\frac{2}{9}\)
\(\Leftrightarrow\frac{1}{3.7}+\frac{1}{4.7}+\frac{1}{4.9}+...+\frac{2}{n}.\left(n+1\right)=\frac{2}{9}\)
\(\Leftrightarrow\frac{2}{2.3.7}+\frac{2}{2.4.7}+\frac{2}{2.4.9}+...+\frac{2}{n}.\left(n+1\right)=\frac{2}{9}\)
\(\Leftrightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{n}.\left(n+1\right)=\frac{2}{9}\)
\(\Leftrightarrow2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{n}-\frac{1}{n}+1\right)=\frac{2}{9}\)
\(\Leftrightarrow2.\left(\frac{1}{6}-\frac{1}{n}+1\right)=\frac{2}{9}\)
\(\Leftrightarrow\frac{1}{6}-\frac{1}{n}+1=\frac{1}{9}\)
\(\Leftrightarrow\frac{1}{n}+1=\frac{1}{6}-\frac{1}{9}\)
\(\Leftrightarrow\frac{1}{n}+1=\frac{1}{18}\)
\(\Leftrightarrow n+1=18\)
\(\Leftrightarrow n=17\)
Vậy \(n=17\)
