Question

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

Answer 1

=\(3\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)

\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)\)\(=3\left(\frac{1}{20}-\frac{1}{80}\right)\)

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

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

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

Answer 2

Katherine Lilly Filbert đúng rồi

Answer 3

1/3=3/20*23+3/23*26+...+3/77+80

1/3=1/20-1/23+1/23-1/26+...+1/77-1/80

1/3=1/20-1/80

1/3=3/80

-> 3/3=3/80*3

->9/80

Vì 9/80<1 nên: => 3^2/20*23+3^2/23*26+...+3^2/77*80