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

Cho em mượn acc bang bang

em có làm được bài này ko?

What ?

Ta có \(\frac{1}{1945^2}+\frac{1}{1946^2}+\frac{1}{1947^2}+...+\frac{1}{1975^2}\)

\(< \frac{1}{1944\cdot1945}+\frac{1}{1945\cdot1946}+...+\frac{1}{1974.1975}\)

\(=\frac{1}{1944}-\frac{1}{1945}+\frac{1}{1945}-\frac{1}{1946}+...+\frac{1}{1974}-\frac{1}{1975}\)

=\(\frac{1}{1944}-\frac{1}{1975}< \frac{1}{1944}\)

\(\Rightarrow\frac{1}{1945^2}+\frac{1}{1946^2}+\frac{1}{1947^2}+..+\frac{1}{1975^2}< \frac{1}{1944}\)

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1/1945² < 1/ 1944.1945

1/1946² < 1/ 1945.1946

....

1/1975² < 1/ 1974.1975

=> 1/11945² +1/11946^2+....+1/11975^{2 <} 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

Ta có : P = \(\dfrac{1}{1975}\left(\dfrac{2}{1945}-1\right)-\dfrac{1}{1945}\left(1-\dfrac{2}{1975}\right)+\dfrac{1974}{1975}.\dfrac{1946}{1945}\)

\(-\dfrac{3}{1975.1945}\)

= \(\dfrac{2}{1975.1945}-\dfrac{1}{1975}-\dfrac{1}{1945}+\dfrac{2}{1975.1945}+\dfrac{1974}{1975}.\dfrac{1946}{1945}\)

\(-\dfrac{3}{1975.1945}\)

= \(\dfrac{2+2+1974.1946-3-1975-1945}{1975.1945}\)

= \(\dfrac{2+2+1974.1946-3-1975-1945}{1975.1945}\)

= \(\dfrac{1973}{1975}\)

E hèm