Question

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

CMR

\(\dfrac{7}{12}< A< \dfrac{5}{6}\)

Answer 1

A= \(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

= \(1-\dfrac{1}{100}\)

= \(\dfrac{99}{100}\)

Ta có :

\(\dfrac{7}{12}=\dfrac{7\cdot50}{12\cdot50}=\dfrac{350}{600}\)

\(\dfrac{99}{100}=\dfrac{99\cdot6}{100\cdot6}=\dfrac{594}{600}\)

\(\dfrac{5}{6}=\dfrac{5\cdot100}{6\cdot100}=\dfrac{500}{600}\)

Chỗ này hình như bn viết sai đề nha

QĐMS lên phải là \(\dfrac{7}{12}< \dfrac{5}{6}< A\) chứ

Bn xem lại đề ik nha ☺

THANKSSSSSSSSSSSS