Question

Tính giá trị biểu thức: A=(1+\(\frac{2}{1.4}\))(1+\(\frac{2}{2.5}\))(1+\(\frac{2}{3.6}\))(1+\(\frac{2}{4.7}\))...(1+\(\frac{2}{2019.2022}\))

Answer 1

+ \(n\left(n+3\right)+2\) \(=n^2+3n+2\)

\(=n^2+2n+n+2=\left(n+1\right)\left(n+2\right)\)

\(A=\frac{1\cdot4+2}{1\cdot4}\cdot\frac{2\cdot5+2}{2\cdot5}\cdot...\cdot\frac{2019\cdot2022+2}{2019\cdot2022}\)

\(=\frac{2\cdot3}{1\cdot4}\cdot\frac{3\cdot4}{2\cdot5}\cdot...\cdot\frac{2020\cdot2021}{2019\cdot2022}\)

\(=\frac{2\cdot3\cdot..\cdot2020}{1\cdot2\cdot...\cdot2019}\cdot\frac{3\cdot4\cdot...\cdot2021}{4\cdot5\cdot...\cdot2022}\)

\(=2020\cdot\frac{3}{2022}=\frac{1010}{337}\)