Question

Chứng minh rằng:

\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+...+\dfrac{1}{97.99}+\dfrac{1}{98.100}\)<\(\dfrac{3}{4}\)

Giúp mình nhé

Answer 1

Đặt A=\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{98.100}\)

A=\(\left(\dfrac{1}{1.3}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{1}{2.4}+...+\dfrac{1}{98.100}\right)\)

A=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right)+\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)

A=\(\dfrac{98}{99}-\dfrac{49}{100}\)

A=\(\dfrac{4949}{9900}\)

Mà \(\dfrac{3}{4}=\dfrac{7425}{9900}\)

Vậy A<\(\dfrac{3}{4}\)

Answer 2

Bạn hãy tính \(\dfrac{1}{1.3}+...+\dfrac{1}{98.100}\)= \(\dfrac{4949}{9900}\) sau đo chỉ cần chứng minh nó nhỏ hơn bằng cách quy đồng .