Question

Các bn làm nhanh lên nhé <3

Answer 1

Ta có \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{20^2}>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{20.21}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{20}-\dfrac{1}{21}=\dfrac{1}{2}-\dfrac{1}{21}=\dfrac{19}{42}>\dfrac{18}{42}=\dfrac{3}{7}\left(đpcm\right)\)