Cho A=1/1^2+1/2^2+1/3^2+1/4^2+....+1/50^2
chứng tỏ rằng A<2
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CHo A=1/3^2+1/4^2+1/5^2+...+1/50^2. Chứng tỏ rằng 1/4<A<4/9
Cho A=1+1/2+1/3+1/4+...+1/2^100-1.Chứng tỏ rằng 50<A<100
M=1/2^2+1/3^2+1/4^2+...+1/2021^2
chứng minh rằng:1/3<M<1
`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`
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+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: \(\dfrac{1}{3}< M< 1\)
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Giải:
\(M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}\)
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{2021^2}=\dfrac{1}{2021.2021}< \dfrac{1}{2020.2021}\)
\(\Rightarrow M< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020.2021}\)
\(\Rightarrow M< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)
\(\Rightarrow M< \dfrac{1}{1}-\dfrac{1}{2021}< 1\)
\(\Rightarrow M< 1\left(1\right)\)
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\)
...
\(\dfrac{1}{2021^2}=\dfrac{1}{2021.2021}>\dfrac{1}{2021.2022}\)
\(\Rightarrow M>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2021.2022}\)
\(\Rightarrow M>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(\Rightarrow M>\dfrac{1}{2}-\dfrac{1}{2022}=\dfrac{505}{1011}=\dfrac{1}{3}+\dfrac{56}{337}>\dfrac{1}{3}\left(2\right)\)
Vậy \(\dfrac{1}{3}< M< 1\) (đpcm)
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chứng minh rằng số A(n) = 2^3n +1 chia hết cho 3^(n+1) nhưng không chia hết cho 3^(n+2chứng minh rằng số A(n) = 2^3^n +1 chia hết cho 3^(n+1) nhưng không chia hết cho 3^(n+2)
a) A= 1/12+1/22+1/32+1/42+1/52+...+1/502 chứng tỏ A < 2
b) B= 21+22+23+24+...+230 chứng tỏ rằng B chia hết cho 21
a) \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+1-\frac{1}{50}\)
\(=2-\frac{1}{50}< 2\)
\(\Rightarrow A< 2\)
b) Ta thấy : 21 = 3 .7 ( 3 ; 7 ) = 1
để chứng minh B \(⋮\)21 , ta cần chứng minh B \(⋮\)3 và 7
Ta có :
B = 21 + 22 + 23 + 24 + ... + 230
B = ( 2 + 22 ) + ( 23 + 24 ) + ... + ( 229 + 230 )
B = 2 . ( 1 + 2 ) + 23 . ( 1 + 2 ) + ... + 229 . ( 1 + 2 )
B = 2 . 3 + 23 . 3 + ... + 229 . 3
B = ( 2 + 23 + ... + 229 ) . 3 \(⋮\)3 ( 1 )
Lại có : B = 21 + 22 + 23 + 24 + ... + 230
B = ( 21 + 22 + 23 ) + ( 24 + 25 + 26 ) + ... + ( 228 + 229 + 230 )
B = 2 . ( 1 + 2 + 22 ) + 24 . ( 1 + 2 + 22 ) + ... + 228 . ( 1 + 2 + 22 )
B = 2 . 7 + 24 . 7 + ... + 228 . 7
B = ( 2 + 24 + ... + 228 ) . 7 \(⋮\)7 ( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\)B \(⋮\)21
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Chứng tỏ rằng:
a,1/(1*2)+1/(2*3)+1/(3*4)+...+1/(49*50)<1
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)\(<1\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}<1\)
Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}<1\)
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\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}<1\)
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cho A = 1 -1/2 + 1/3 - 1/4 +....+1/49 - 1/50. chứng tỏ rằng 7/12<A
A=1 - 1/2 + 1/3 - 1/4 +..+ 1/49 - 1/50
A= 1-( 1/2 + 1/3 ) - ( 1/4 + 1/5 ) -.....-(1/48 + 1/49) - 1/50
A=1 - 5/6 - 9/20 -.....-97/2352 - /150
A= 1 -............cho con lai tu lam nha
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Cho a=1×2×3×4.......49×50.Chứng tỏ rằng các số sau đây ld hợp số :a+2;a+3;.......;a+50