chứng tỏ rằng: \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{60^2}< \dfrac{1}{9}\)

chứng tỏ rằng:\(\dfrac{1}{2^3}+\dfrac{1}{4^2}+...+\dfrac{1}{60^2}< \dfrac{1}{9}\)

I\(\dfrac{3}{4}\)x-\(\dfrac{1}{2}\)I -1=\(\dfrac{1}{4}\)