Cho A=1/2^2+1/3^2+1/4^2+...+1/100^2
chứng minh A<3/4
K
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PH
0
PH
0
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: \(\dfrac{1}{3}< M< 1\)
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
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\\ \Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=4\\ \Rightarrow2+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=4\\ \Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\\ \Rightarrow\dfrac{a+b+c}{abc}=1\\ \Rightarrow a+b+c=abc\left(dpcm\right)\)
B
0
14 tháng 11 2023
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
VT
0
29 tháng 6 2016
\(a.A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(2A-A=1-\frac{1}{2^{99}}\)
\(A=1-\frac{1}{2^{99}}< 1\)
\(b.B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{203}{3^{100}}< 3\)
\(A< \frac{3}{4}\)
Ủng hộ mk nha ^_^