\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2021.2023}\right)\)
\(=\dfrac{4}{1.3}.\dfrac{9}{2.4}...\dfrac{4088484}{2021.2023}\)
\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{2022.2022}{2021.2023}\)
\(=\dfrac{2.2022}{1.2023}\)