Chứng tỏ rằng 1/1945^2+1?1946^2+...+1/1975^2<1/1944. Ai pít thì giúp mình ná
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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
\(\dfrac{1}{1945^2}< \dfrac{1}{1944^2}\\ \dfrac{1}{1946^2}< \dfrac{1}{1944^2}\\ \dfrac{1}{1947^2}< \dfrac{1}{1944^2}\\ ...\\ \dfrac{1}{1975^2}< \dfrac{1}{1944^2}\\ \Leftrightarrow\dfrac{1}{1945^2}+\dfrac{1}{1946^2}+\dfrac{1}{1947^2}+...+\dfrac{1}{1975^2}< \dfrac{1}{1944^2}+\dfrac{1}{1944^2}+\dfrac{1}{1944^2}+...+\dfrac{1}{1944^2}\left(31\text{ số }\dfrac{1}{1944^2}\right)=31\cdot\dfrac{1}{1944^2}< 1944\cdot\dfrac{1}{1944^2}=\dfrac{1}{1944}\)
Vậy \(\dfrac{1}{1945^2}+\dfrac{1}{1946^2}+\dfrac{1}{1947^2}+...+\dfrac{1}{1975^2}< \dfrac{1}{1944}\)
Ta có : P = \(\dfrac{1}{1975}\left(\dfrac{2}{1945}-1\right)-\dfrac{1}{1945}\left(1-\dfrac{2}{1975}\right)+\dfrac{1974}{1975}.\dfrac{1946}{1945}\)
\(-\dfrac{3}{1975.1945}\)
= \(\dfrac{2}{1975.1945}-\dfrac{1}{1975}-\dfrac{1}{1945}+\dfrac{2}{1975.1945}+\dfrac{1974}{1975}.\dfrac{1946}{1945}\)
\(-\dfrac{3}{1975.1945}\)
= \(\dfrac{2+2+1974.1946-3-1975-1945}{1975.1945}\)
= \(\dfrac{2+2+1974.1946-3-1975-1945}{1975.1945}\)
= \(\dfrac{1973}{1975}\)
Ta có \(\frac{1}{1945^2}+\frac{1}{1946^2}+\frac{1}{1947^2}+...+\frac{1}{1975^2}\)
\(< \frac{1}{1944\cdot1945}+\frac{1}{1945\cdot1946}+...+\frac{1}{1974.1975}\)
\(=\frac{1}{1944}-\frac{1}{1945}+\frac{1}{1945}-\frac{1}{1946}+...+\frac{1}{1974}-\frac{1}{1975}\)
=\(\frac{1}{1944}-\frac{1}{1975}< \frac{1}{1944}\)
\(\Rightarrow\frac{1}{1945^2}+\frac{1}{1946^2}+\frac{1}{1947^2}+..+\frac{1}{1975^2}< \frac{1}{1944}\)