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 \(\frac{1}{200}< A< \frac{1}{99}\)
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