Ta có:
\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}>\frac{1}{25}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)
\(=\frac{1}{25}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{25}+\frac{1}{6}-\frac{1}{101}>\frac{1}{6}+\frac{1}{25}-\frac{1}{100}=\frac{1}{6}+\frac{3}{100}>\frac{1}{6}\left(1\right)\)
\(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\left(2\right)\)
Từ (1) và (2) suy ra:\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(đpcm\right)\)
đạt 1/52+.........+1/1002=S
1/52>1/5*6
.....................
1/1002>1/100*101
=>S>1/5*6+.............+1/100*101=1/5-1/6+....+1/100-1/101=1/5-1/101=96/505>96/576=1/6
vậ S>1/6
1/52<1/4*5
.....................
1/1002<1/99*100
=>S<1/4*5+................+1/99*100=1/4-1/5+.....+1/99-1/100=1/4-1/100=6/25<6/24=1/4
Vậy 1/6<S<1/4
Bạn so sánh
\(\frac{1}{4.5}>\frac{1}{5^2}>\frac{1}{5.6}\)
\(\frac{1}{5.6}>\frac{1}{6^2}>\frac{1}{6.7}\)
.......
\(\frac{1}{99.100}>\frac{1}{100^2}>\frac{1}{100.101}\)
Ta gọi \(\frac{1}{5.6}+\frac{1}{6.7}+..+\frac{1}{100.101}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
Mà \(\frac{1}{5.6}=\frac{1}{5}-\frac{1}{6}\)
=> \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}< \frac{1}{5^2}+\frac{1}{6^2}+..+\frac{1}{100^2}< \frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{5}-\frac{1}{101}< \frac{1}{5^2}+\frac{1}{6^2}+..+\frac{1}{100^2}< \frac{1}{4}-\frac{1}{100}\)
Mà \(\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\) <=> \(\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
Từ đó ta có điều phải chứng minh!
\(\text{Ta có : }\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
\(>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}\)
\(=\frac{96}{505}>\frac{96}{576}=\frac{1}{6}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}>\frac{1}{6}\left(1\right)\)
\(\text{Lại có:}\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
\(< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(2\right)\)
\(\text{Từ (1) và (2) ta có :}\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(\text{đpcm}\right)\)