Question

\(\frac{1}{20x23}\)+\(\frac{1}{23x26}\)+...+\(\frac{1}{77x80}\)< \(\frac{1}{9}\)

Answer 1

\(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+...+\frac{1}{77\cdot80}\)

\(=\frac{1}{3}\left[\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+...+\frac{3}{77\cdot80}\right]\)

\(=\frac{1}{3}\left[\frac{1}{20}-\frac{1}{23}+...+\frac{1}{77}-\frac{1}{80}\right]\)

\(=\frac{1}{3}\left[\frac{1}{20}-\frac{1}{80}\right]\)

\(=\frac{1}{3}\left[\frac{4}{80}-\frac{1}{80}\right]\)

\(=\frac{1}{3}\cdot\frac{3}{80}=\frac{1}{1}\cdot\frac{1}{80}=\frac{1}{80}\)

Mà \(\frac{1}{80}< \frac{1}{9}\)nên \(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+...+\frac{1}{77\cdot80}< \frac{1}{9}\)

Vậy : ...

Answer 2

\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\)

\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)

\(=\frac{1}{3}.\frac{3}{80}\)

\(=\frac{1}{80}< \frac{1}{9}\)