Question

Chứng minh rằng              \(\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}< 1\)

Answer 1

Ta có
\(A=\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}\)
\(A=3^2\left(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\right)\)
\(A=3^2\cdot\frac{1}{3}\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(A=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(A=3\cdot\frac{3}{80}=\frac{9}{80}< 1\left(9< 80\right)\)