Cho
A = 1/1^2+1/2^2+1/3^2+1/4^2+..+1/50^2
Chứng minh A < 2.
K
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TN
4
5 tháng 5 2017
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}< 1< 2\Rightarrow A< 2\Rightarrowđpcm\)
17 tháng 5 2021
`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`
17 tháng 5 2021
+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: \(\dfrac{1}{3}< M< 1\)
TN
8 tháng 4 2016
đặt B=1/2.3+1/3.4+...+1/49.50
=1/1.2+1/2.3+1/3.4+...+1/49.50
=1-1/2+1/2-1/3+...+1/49-1/50
=1-1/50<1 (1)
Mà 1<2(2)
A =1/1+1/2.2+1/3.3+...+1/50.50<1-1/2+1/2-1/3+...+1/49-1/50 (3)
từ (1),(2),(3) =>A<2
BM
8 tháng 4 2016
Ta có : \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...........+\frac{1}{50^2}=1+\frac{1}{2^2}+........+\frac{1}{50^2}\)
=> \(A<1+\frac{1}{1.2}+\frac{1}{2.3}+.............+\frac{1}{49.50}\)
=> \(A<1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{49}-\frac{1}{50}\)
=> \(A<2-\frac{1}{50}\Rightarrow A<2\)
Vậy A nhỏ hơn 2
PH
0
PH
0
HT
3
R
26 tháng 11 2021
Ta có :
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=1a^2+1b^2+1c^2+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}\)
\(=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)
\(=2^2=2=2+2.\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)
\(=\dfrac{c}{abc}+\dfrac{a}{abc}+\dfrac{b}{abc}=\dfrac{abc}{abc}\)
\(=a+b+c\)
\(=abc\)
26 tháng 11 2021
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\\ \Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=4\\ \Rightarrow2+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=4\\ \Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\\ \Rightarrow\dfrac{a+b+c}{abc}=1\\ \Rightarrow a+b+c=abc\left(dpcm\right)\)
dễ lắm bn ạ nhưng mk ko bt lm hhh