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
@tranlethanhtruc: Bn ấy lm sai.