Question

chứng minh rằng với mọi n thuộc N và n>=2 thì

\(\frac{3}{9.14}+\frac{3}{14.19}+\frac{3}{19.24}+...+\frac{3}{\left(5n+1\right)\left(5n+4\right)}< \frac{1}{15}\)

Answer 1

\(\frac{3}{9.14}+\frac{3}{14.19}+\frac{3}{19.24}+....+\frac{3}{\left(5n+1\right)\left(5n+4\right)}\)

\(=\frac{3}{5}\left(\frac{5}{9.14}+\frac{5}{14.19}+\frac{5}{19.24}+....+\frac{5}{\left(5n+1\right)\left(5n+4\right)}\right)\)

\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+....+\frac{1}{5n+1}-\frac{1}{5n+4}\right)\)

\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{5n+4}\right)\)

\(=\frac{1}{15}-\frac{3}{5\left(5n+4\right)}< \frac{1}{15}\) (đpcm)