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Ta có : A = \(\frac{1}{100^2}+\frac{1}{101^2}+...+\frac{1}{199^2}=\frac{1}{100.100}+\frac{1}{101.101}+...+\frac{1}{199.199}\)
> \(\frac{1}{100.101}+\frac{1}{101.102}+...+\frac{1}{199.200}=\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+...+\frac{1}{199}-\frac{1}{200}\)
= \(\frac{1}{100}-\frac{1}{200}=\frac{1}{200}\Rightarrow A>\frac{1}{200}\left(1\right)\)
Lại có : A = \(\frac{1}{100^2}+\frac{1}{101^2}+...+\frac{1}{199^2}=\frac{1}{100.100}+\frac{1}{101.101}+...+\frac{1}{199.199}\)
\(< \frac{1}{99.100}+\frac{1}{100.101}+...+\frac{1}{198.199}=\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}+...+\frac{1}{198}-\frac{1}{199}\)
\(=\frac{1}{99}-\frac{1}{199}\Rightarrow A< \frac{1}{99}\left(2\right)\)
Từ (1) và (2) => \(\frac{1}{200}< A< \frac{1}{99}\left(\text{ĐPCM}\right)\)
Cho A=\(\frac{1}{100^2}+\frac{1}{101^2}+......................+\frac{1}{198^2}+\frac{1}{199^2}\)
CMR:\(\frac{1}{200}< A< \frac{1}{99}\)
+)Ta có:A=\(\frac{1}{100^2}+\frac{1}{101^2}+......................+\frac{1}{198^2}+\frac{1}{199^2}\)
=>A=\(\frac{1}{100.100}+\frac{1}{101.101}+...........+\frac{1}{198.198}+\frac{1}{199.199}\)
+)Ta thấy :\(\frac{1}{100.100}\)>\(\frac{1}{100.101}\)
\(\frac{1}{101.101}>\frac{1}{101.102}\)
.............................................
\(\frac{1}{198.198}>\frac{1}{198.199}\)
\(\frac{1}{199.199}>\frac{1}{199.200}\)
=> \(\frac{1}{100.100}+\frac{1}{101.101}+...........+\frac{1}{198.198}+\frac{1}{199.199}\)>\(\frac{1}{100.101}+\frac{1}{101.102}+................+\frac{1}{198.199}+\frac{1}{199.200}\)
=>A>\(\frac{1}{100.101}+\frac{1}{101.102}+................+\frac{1}{198.199}+\frac{1}{199.200}\)
=>A>\(\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+........+\frac{1}{198}-\frac{1}{199}+\frac{1}{199}-\frac{1}{200}\)
=>A>\(\frac{1}{100}-\frac{1}{200}=\frac{2}{200}-\frac{1}{200}=\frac{1}{200}\)
=>A>\(\frac{1}{200}\)(1)
+)Ta lại có:
A=\(\frac{1}{100^2}+\frac{1}{101^2}+......................+\frac{1}{198^2}+\frac{1}{199^2}\)
=>A=\(\frac{1}{100.100}+\frac{1}{101.101}+...........+\frac{1}{198.198}+\frac{1}{199.199}\)
+)Ta lại thấy:\(\frac{1}{100.100}< \frac{1}{99.100}\)
\(\frac{1}{101.101}< \frac{1}{100.101}\)
................................................
\(\frac{1}{198.198}< \frac{1}{197.198}\)
\(\frac{1}{199.199}< \frac{1}{198.199}\)
=>\(\frac{1}{100.100}+\frac{1}{101.101}+...........+\frac{1}{198.198}+\frac{1}{199.199}\)<\(\frac{1}{99.100}+\frac{1}{100.101}+.............+\frac{1}{197.198}+\frac{1}{198.199}\)
=>A<\(\frac{1}{99.100}+\frac{1}{100.101}+.............+\frac{1}{197.198}+\frac{1}{198.199}\)
=>A<\(\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}+...........+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\)
=>A<\(\frac{1}{99}-\frac{1}{199}\)
Mà A<\(\frac{1}{99}-\frac{1}{199}\)
=>A<\(\frac{1}{99}\)(2)
+)Từ (1) và (2)
=>\(\frac{1}{200}< A< \frac{1}{99}\)(ĐPCM)
Vậy \(\frac{1}{200}< A< \frac{1}{99}\)
Chúc bn học tốt
a) Mỗi biểu thức M và N đều có 50 thừa số
Ta thấy \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
Vậy \(M< N\)
b) \(M.N=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{99}{100}.\frac{100}{101}\)
\(=\frac{1}{101}\)
c) Vì \(M< N\)nên \(M.M< M.N\)hay \(M.M< \frac{1}{101}< \frac{1}{100}\). Do đó \(M.M< \frac{1}{100}=\frac{1}{10}.\frac{1}{10}\)suy ra \(M< \frac{1}{10}\)( Vì \(M>0\))
`Answer:`
Tổng: `(200-100):1+1=100` số hạng
Ta có:
\(\frac{1}{101}>\frac{1}{200}\)
\(\frac{1}{102}>\frac{1}{200}\)
...
\(\frac{1}{200}=\frac{1}{200}\)
\(\Rightarrow A>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\)
\(\Rightarrow A>\frac{100}{200}\)
\(\Rightarrow A>\frac{1}{2}\)
Ta có :
1002 > 99 . 100
1012 > 100 . 101
..............
2002 > 199. 200
=> A < \(\frac{1}{99.100}+\frac{1}{100.101}+...+\frac{1}{199.200}=\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}+...+\frac{1}{199}-\frac{1}{200}\)
=> A < \(\frac{1}{99}-\frac{1}{200}< \frac{1}{99}\) \(\left(1\right)\)
Tương tự ta có :
A > \(\frac{1}{100.101}+\frac{1}{101.102}+...+\frac{1}{200.201}\)
=> A > \(\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+...+\frac{1}{200}-\frac{1}{201}\)
=> A > \(\frac{1}{100}-\frac{1}{201}>\frac{1}{100}-\frac{1}{200}\)
=> A > \(\frac{1}{200}\) \(\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\)Ta có :
\(\frac{1}{200}< A< \frac{1}{99}\)
=> ĐPCM