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Đặt A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\)
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2},\dfrac{1}{3^2}< \dfrac{1}{2.3},...,\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(A\)<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
A<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A<\(1-\dfrac{1}{100}=\dfrac{99}{100}\)(đpcm)
Ta có: \(\dfrac{1}{2^2}>\dfrac{1}{2.3},\dfrac{1}{3^2}>\dfrac{1}{3.4},...,\dfrac{1}{100^2}>\dfrac{1}{100.101}\)
A>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\)
A>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
A>\(\dfrac{1}{2}-\dfrac{1}{101}=\dfrac{99}{202}\)(đpcm)
Vậy \(\dfrac{99}{100}>A>\dfrac{99}{202}\)
Ta có \(\frac{1}{1945^2}+\frac{1}{1946^2}+\frac{1}{1947^2}+...+\frac{1}{1975^2}\)
\(< \frac{1}{1944\cdot1945}+\frac{1}{1945\cdot1946}+...+\frac{1}{1974.1975}\)
\(=\frac{1}{1944}-\frac{1}{1945}+\frac{1}{1945}-\frac{1}{1946}+...+\frac{1}{1974}-\frac{1}{1975}\)
=\(\frac{1}{1944}-\frac{1}{1975}< \frac{1}{1944}\)
\(\Rightarrow\frac{1}{1945^2}+\frac{1}{1946^2}+\frac{1}{1947^2}+..+\frac{1}{1975^2}< \frac{1}{1944}\)
Đặt A=1/22+1/32+...+1/1002.Ta có:
A>1/2.3+1/3.4+...+1/100.101=1/2-1/101=99/202
A< 1/1.2+1/2.3+...+1/99.100=1-1/100=99/100
thanks nhìu nha