Question

CM

\(A=\frac{2}{3^2}+\frac{2}{5^2}+...+\frac{2}{2007^2}\) <\(\frac{1003}{2008}\)

\(B=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2006^2}<\frac{334}{2007}\)

\(S=\frac{1}{5^2}+\frac{1}{9^2}+...+\frac{1}{409^2}<\frac{1}{12}\)

Answer 1

Đặt B=\(\frac{2}{4^2}+\frac{2}{6^2}+\frac{2}{8^2}+....+\frac{2}{2008^2}\)

=> A+B= 2\(\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2007^2}+\frac{1}{2008^2}\right)\) <2   \(\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{2006\cdot2007}+\frac{1}{2007\cdot2008}\right)\)

=2\(\left(\frac{1}{2}-\frac{1}{2008}\right)\)=\(\frac{2006}{2008}\)

mà A<B=>A+A<A+B=2006/2008

=>A<1003/2008

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