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Có : 1/1945^2 + 1/1946^2 + ...... + 1/1975^2
< 1/1944.1945 + 1/1945.1946 + ...... + 1/1974.1975
= 1/1944 - 1/1945 +1/1945 - 1/1946 + ...... + 1/1974 - 1/1975
= 1/1944 - 1/1975
< 1/1944
Tk mk nha
Ta có \(\frac{1}{1945^2}+\frac{1}{1946^2}+\frac{1}{1947^2}+...+\frac{1}{1975^2}\)
\(< \frac{1}{1944\cdot1945}+\frac{1}{1945\cdot1946}+...+\frac{1}{1974.1975}\)
\(=\frac{1}{1944}-\frac{1}{1945}+\frac{1}{1945}-\frac{1}{1946}+...+\frac{1}{1974}-\frac{1}{1975}\)
=\(\frac{1}{1944}-\frac{1}{1975}< \frac{1}{1944}\)
\(\Rightarrow\frac{1}{1945^2}+\frac{1}{1946^2}+\frac{1}{1947^2}+..+\frac{1}{1975^2}< \frac{1}{1944}\)
\(\dfrac{1}{1945^2}< \dfrac{1}{1944^2}\\ \dfrac{1}{1946^2}< \dfrac{1}{1944^2}\\ \dfrac{1}{1947^2}< \dfrac{1}{1944^2}\\ ...\\ \dfrac{1}{1975^2}< \dfrac{1}{1944^2}\\ \Leftrightarrow\dfrac{1}{1945^2}+\dfrac{1}{1946^2}+\dfrac{1}{1947^2}+...+\dfrac{1}{1975^2}< \dfrac{1}{1944^2}+\dfrac{1}{1944^2}+\dfrac{1}{1944^2}+...+\dfrac{1}{1944^2}\left(31\text{ số }\dfrac{1}{1944^2}\right)=31\cdot\dfrac{1}{1944^2}< 1944\cdot\dfrac{1}{1944^2}=\dfrac{1}{1944}\)
Vậy \(\dfrac{1}{1945^2}+\dfrac{1}{1946^2}+\dfrac{1}{1947^2}+...+\dfrac{1}{1975^2}< \dfrac{1}{1944}\)
1/19452 < 1/ 1944.1945
1/19462 < 1/ 1945.1946
....
1/19752 < 1/ 1974.1975
=> 1/119452 +1/119462+....+1/119752 < 1/ 1944.1945+1/ 1945.1946+..+1/ 1974.1975=1/1944-1/1945+1/1945-1/1946+....+1/1974-1/1975
=1/19444-1/1975<1/1944