Question

SO SÁNH

Đặt A= \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{39}{19^2-20^2}với1\)

Answer 1

\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{39}{19^2.20^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{39}{361.400}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{361}-\dfrac{1}{400}\)

\(=1-\dfrac{1}{400}< 1\)

Vậy A < 1