Ta có:\(\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)

        \(=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)

        \(=\frac{1}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)

        \(=\frac{1}{5}.\left(1-\frac{1}{5n+6}\right)\)

        \(=\frac{1}{5}.\left(\frac{5n+5}{5n+6}\right)=\frac{n+1}{5n+6}\left(\text{đ}pcm\right)\)

\(A=\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+...+\frac{1}{\left(5n+1\right)\cdot\left(5n+6\right)}\)\(5A=\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{\left(5n+1\right)\cdot\left(5n+6\right)}\)\(5A=\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\)

C = 1/1 . 6 + 1/6 . 11 + 1/11 . 16 + ...+ 1/( 5n + 1 ) . ( 5n + 6 ) 

C = 1/5 . ( 5/1 . 6 + 5/6 . 11 + 5/11 . 16 + ...+ 5/( 5n + 1 ) . ( 5n + 6 )  ) 

C = 1/5 . ( 1 - 1/6 + 1/6 - 1/11 + 1/11 - 1/16 + ...+ 1/5n + 1 - 1/5n + 6  ) 

C = 1/5 . ( 1 - 1/5n + 6 ) 

C = 1/5 . 1 - 1/5 . 1/5n + 6

C = 1/5 - 1/ 5 . ( 5n + 6 ) 

Ta có:

\(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)

\(=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{5n+6}\right)=\frac{1}{5}\left(\frac{5n+6}{5n+6}-\frac{1}{5n+6}\right)=\frac{1}{5}.\frac{5n+5}{5n+6}=\frac{1}{5}.\frac{5\left(n+1\right)}{5n+6}=\frac{5\left(n+1\right)}{5\left(5n+6\right)}=\frac{n+1}{5n+6}\)(ĐPCM)

bạn Phạm Thiết Tường ơi ch mình hỏi sao lại nhân \(\frac{1}{5}\)với \(\frac{1}{1}-\frac{1}{5n+6}\)vậy

\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)

\(=\dfrac{1}{5}\cdot\dfrac{5n+6-1}{5n+6}\)

\(=\dfrac{n+1}{5n+6}=VP\)