Question

Cho A=\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{99\times100}\)

So sánh A với 1

Answer 1

Đặt A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

=> A = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

=> A = 1 - \(\dfrac{1}{100}\) = \(\dfrac{99}{100}\)

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

=> A < 1

Answer 2

A =

=> A =

=> A = 1 - =

=> 1 =

=> A < 1