Question

CMR : E = \(1-\frac{1}{2^2}-\frac{1}{3^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)

F = \(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{200^2}< \frac{1}{2}\)

H = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)

Answer 1

\(E=1-\frac{1}{2^2}-\frac{1}{3^2}-..........-\frac{1}{2004^2}\)

\(E=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+..........+\frac{1}{2014^2}\right)\)

Ta có : \(E< 1-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{2003.2004}\right)\\ \)

Đặt A= \(1-\left(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2003.2004}\right)\\ =>A=1-\left(1-\frac{1}{2004}\right)\\ =>A=1-\frac{2003}{2004}\\ =>A=\frac{1}{2004}\)

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