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`M=1/2^2+1/3^2+1/4^2+...+1/2021^2`
Vì `1/2^2>1/(2.3)`
`1/(3^2)>1/(3.4)`
`....................`
`1/2021^2>1/(2021.2022)`
`=>M>1/(2.3)+1/(3.4)+............+1/(2021.2022)`
`=>M>1/2-1/3+1/3-1/4+..........+1/2021-1/2022`
`=>M>1/2-1/2022=505/1011=1/3+56/337>1/3(1)`
Vì `1/2^2<1/(1.2)`
`1/(3^2)<1/(2.3)`
`....................`
`1/2021^2<1/(2021.2020)`
`=>M<1/(1.2)+1/(2.3)+............+1/(2020.2021)`
`=>M<1-1/2+1/2-1/3+..........+1/2020-1/2021`
`=>M<1-1/2021<1(2)`
`(1)(2)=>1/3<M<1`
+Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3};\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4};\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5};...;\dfrac{1}{2021^2}=\dfrac{1}{2021.2021}>\dfrac{1}{2021.2022}\)\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2021.2022}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2021}-\dfrac{1}{2022}=\dfrac{1}{2}-\dfrac{1}{2022}=\dfrac{505}{1011}>\dfrac{1}{3}\left(1\right)\)+Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{2021^2}< \dfrac{1}{2020.2021}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020.2021}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}=1-\dfrac{1}{2021}< 1\left(2\right)\)Từ (1) và (2) suy ra: \(\dfrac{1}{3}< M< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ai chia buon voi toi voi nich o thu 9 cua tui bi khoa den ngay mai roi huhu
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow A<1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{49.50}\)
\(\Rightarrow A<1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow A<1+\left(1-\frac{1}{50}\right)\)
\(\Rightarrow A<1+\frac{49}{50}\)
\(\Rightarrow A<\frac{99}{50}\)
Vì \(\frac{99}{50}<2=\frac{100}{50}\Rightarrow A<2\) ĐPCM
Ta có:
\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};......;\frac{1}{50^2}<\frac{1}{49.50}\)
Do đó \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}<1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
\(\Rightarrow A<1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}<2\)
=>A<2(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: A < \(\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
Lại có: \(\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+\left(\frac{1}{1}-\frac{1}{50}\right)\)
\(=1+\frac{49}{50}\)
Mà 1+49/50<2 nên A<1+49/50<2
Vậy A<2
đặt B=1/1.2+1/2.3+...+1/49.50
ta có:
A=1/1^2+1/2^2+1/3^2+1/4^2+....+1/50^2<B=1/1.2+1/2.3+...+1/49.50 (1)
B=1/1.2+1/2.3+...+1/49.50
=1-1/2+1/2-1/3+...+1/49-1/50
=1-1/50<1 (2)
từ (1) va (2)=>A<B<2
=>A<2