Question

[Hồ Thu Giang]

Chứng minh

\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+....+\frac{1}{10^2+11^2}<\frac{9}{20}\)

Answer 1

[Đinh Tuấn Việt]

đơn giải thôi nhưng mình ko bấn fx đc

Answer 2

[Khánh Vy]

xét vế trái : \(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{221}\)

ta có : \(T< \frac{1}{5}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{220}\)

             \(=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

             \(=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

              \(=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{11}\right)< \frac{1}{5}+\frac{1}{4}\Rightarrow T< \frac{9}{20}\)