Question

so sánh  \(\frac{1}{101^2}+\frac{1}{102^2}+...+\frac{1}{205^2}v\text{à}\frac{1}{2^2\cdot3\cdot5^2\cdot7}\)

Answer 1

Đặt A = \(\frac{1}{101^2}+\frac{1}{102^2}+...+\frac{1}{205^2}\)

=> A < \(\frac{1}{100.101}+\frac{1}{101.102}+....+\frac{1}{204.205}\)

=> A < \(\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+...+\frac{1}{204}-\frac{1}{205}\)

=> A < \(\frac{1}{100}-\frac{1}{205}\)

=> A < \(\frac{1}{2100}\)

Đặt B = \(\frac{1}{2^2.3.5^2.7}=\frac{1}{2100}\)

=> A < B

=> \(\frac{1}{101^2}+\frac{1}{102^2}+...+\frac{1}{205^2}

Answer 2

giỏi lắm mình cũng biết làm chỉ hỏi chơi thôi 

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