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Ta có : \(C=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow\frac{2}{5}< C\) (1)
Ta có : \(C=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}=1-\frac{1}{9}=\frac{8}{9}\)
\(\Rightarrow C< \frac{8}{9}\) (2)
Từ (1) và (2) ta thấy :
Vậy \(\frac{2}{5}< C< \frac{8}{9}\) ( đpcm )
Ta có: \(\frac{3}{4}=1-\frac{1}{4}=1-\frac{1}{2^2}\); \(\frac{8}{9}=1-\frac{1}{9}=1-\frac{1}{3^2}\)
\(\frac{15}{16}=1-\frac{1}{16}=1-\frac{1}{4^2}\); ...; \(\frac{9999}{10000}=1-\frac{1}{10000}=1-\frac{1}{100^2}\)
=> \(C=\left(1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
=> \(C=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)=99-B\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
=> \(B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)
=> A > 99-1 = 98
=> B > 98
C = \(\dfrac{\dfrac{1}{9}-\dfrac{5}{6}-4}{\dfrac{7}{12}-\dfrac{1}{36}-10}\)
C = \(\dfrac{\dfrac{6-45-216}{54}}{\dfrac{21-1-360}{36}}\)
C = \(\dfrac{\dfrac{-85}{18}}{-\dfrac{85}{9}}\)
C = \(\dfrac{1}{2}\)
C= (1 - \(\frac{1}{2^2}\))+(1 - \(\frac{1}{3^2}\) )+(1 - \(\frac{1}{4^2}\))+.......+(1 - \(\frac{1}{100^2}\))
=98 - (\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+........+\(\frac{1}{100^2}\))
=> C< 98 bn xem lai nha hinh nhu de sai phai cong den \(\frac{9999}{10000}\)
Uk hinh nhu sai