Question

Bài 1: Tính

a, A= (1-\(\dfrac{1}{2}\))*(1-\(\dfrac{1}{3}\))*(1-\(\dfrac{1}{4}\))*...*(1-\(\dfrac{1}{2017}\))

b, B= \(\dfrac{1^2}{1\cdot2}\)*\(\dfrac{2^2}{2\cdot3}\)*\(\dfrac{3^2}{3\cdot4}\)*...*\(\dfrac{99^2}{99\cdot100}\)

Answer 1

Bài 1 :

\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2017}\right)\) \(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2015}{2016}.\dfrac{2016}{2017}=\dfrac{1.2.3.4....2015.2016}{2.3.4.5...2016.2017}=\dfrac{1}{2017}\)

Answer 2

\(B=\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}....\dfrac{99^2}{99.100}\)

\(=\dfrac{1.1}{1.2}.\dfrac{2.2}{2.3}.\dfrac{3.3}{3.4}....\dfrac{99.99}{99.100}=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}=\dfrac{1.2.3...99}{2.3.4...100}=\dfrac{1}{100}\)