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A= 1/100x100+1/101x101+..........+1/199x199
Vì 1/100x100<99x100
1/101x101<100x101
...........
1/199x199 < 1/198x199
=) A< 1/99x100+1/100x101+...+1/198x199
A<1/99-1/100+1/100-1/101+.....+1/198-199
A<100/19701=0,0050....
Mà 1/100=0,01
=> A<1/100
K đúng nhé
Đặt \(S=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{199}-\frac{1}{200}\)
\(S=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(S=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Ta có đpcm
ta thấy : \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};\frac{1}{4^2}>\frac{1}{4.5};...;\frac{1}{199^2}>\frac{1}{199.200}\)
suy ra: \(M>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{199.200}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{199}-\frac{1}{200}=\frac{1}{2}-\frac{1}{200}\)
=\(\frac{100}{200}-\frac{1}{200}=\frac{99}{200}\)
=> \(M>\frac{99}{200}\)
ta cũng thấy: \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};\frac{1}{4^2}<\frac{1}{3.4};...;\frac{1}{199^2}<\frac{1}{198.199}\)
suy ra:\(M<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{198.199}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{198}-\frac{1}{199}=\frac{1}{1}-\frac{1}{199}\)
=\(\frac{199}{199}-\frac{1}{199}=\frac{198}{199}\)
=>\(M<\frac{198}{199}\)
vậy \(\frac{99}{200}
Bài 1:
C = 1/101 + 1/102 + 1/103 + ... + 1/200
Có:
C < 1/101 + 1/101 + 1/101 + ... + 1/101
C < 100 . 1/101
C < 100/101
Mà 100/101 < 1
=> C < 1 (1)
Có:
C > 1/200 + 1/200 + 1/200 + ... + 1/200
C > 100 . 1/200
C > 1/2 (2)
Từ (1) và (2)
=> 1/2<C<1
Ủng hộ nha mk làm tiếp