Question

[TrịnhAnhKiệt]

Cho P=(1+\(\dfrac{1}{1.3}\)).(1+\(\dfrac{1}{2.4}\)).(1+\(\dfrac{1}{3.5}\)). ... .(1+\(\dfrac{1}{2023.2025}\)) là 1 tích của 2023 thừa số
Tính giá trị của P (để dưới dạng phân số tối giản)

Answer 1

[Nguyễn Lê Phước Thịnh]

\(P=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{2023\cdot2025}\right)\)

\(=\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2024^2-1}\right)\)

\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2024^2}{\left(2024-1\right)\left(2024+1\right)}\)

\(=\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot...\cdot\dfrac{2024\cdot2024}{2023\cdot2025}\)

\(=\dfrac{2\cdot3\cdot...\cdot2024}{1\cdot2\cdot3\cdot...\cdot2023}\cdot\dfrac{2\cdot3\cdot...\cdot2024}{3\cdot4\cdot5\cdot...\cdot2025}\)

\(=2024\cdot\dfrac{2}{2025}=\dfrac{4048}{2025}\)