Question

[Tran Le Hoang Vu]

 tinh B

(1+\(\dfrac{1}{1x3}\))x(1+\(\dfrac{1}{2x4}\))+(1+\(\dfrac{1}{3x5}\))+....+(1+\(\dfrac{1}{2022x2024}\))=B

Answer 1

[Nguyễn Lê Phước Thịnh]

Sửa đề: \(B=\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{2022\cdot2024}\right)\)

\(=\left(1+\dfrac{1}{2^2-1}\right)\cdot\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2023^2-1}\right)\)

\(=\dfrac{2^2}{2^2-1}\cdot\dfrac{3^2}{3^2-1}\cdot...\cdot\dfrac{2023^2}{2023^2-1}\)

\(=\dfrac{2\cdot3\cdot...\cdot2023}{1\cdot2\cdot...\cdot2022}\cdot\dfrac{2\cdot3\cdot....\cdot2023}{3\cdot4\cdot...\cdot2024}\)

\(=\dfrac{2023}{1}\cdot\dfrac{2}{2024}=\dfrac{2023}{1012}\)