a: Ta có; \(B=\frac{1^2}{2^2-1}\cdot\frac{3^2}{4^2-1}\cdot\ldots\cdot\frac{2019^2}{2020^2-1}\)
\(=\frac{1^2}{1\cdot3}\cdot\frac{3^2}{3\cdot5}\cdot\ldots\cdot\frac{2019^2}{2019\cdot2021}\)
\(=\frac{1\cdot3\cdot\ldots\cdot2019}{1\cdot3\cdot\ldots\cdot2019}\cdot\frac{1\cdot3\cdot\ldots\cdot2019}{3\cdot5\cdot\ldots\cdot2021}=\frac{1}{2021}\)
b: \(C=\frac{3^2-1}{5^2-1}\cdot\frac{7^2-1}{9^2-1}\cdot\ldots\cdot\frac{99^2-1}{101^2-1}\)
\(=\frac{\left(3-1\right)\left(3+1\right)}{\left(5-1\right)\left(5+1\right)}\cdot\frac{\left(7-1\right)\left(7+1\right)}{\left(9-1\right)\left(9+1\right)}\cdot\ldots\cdot\frac{\left(99-1\right)\left(99+1\right)}{\left(101-1\right)\left(101+1\right)}\)
\(=\frac{2\cdot4}{4\cdot6}\cdot\frac{6\cdot8}{8\cdot10}\cdot\ldots\cdot\frac{98\cdot100}{100\cdot102}\) \(=\frac{2\cdot4\cdot\ldots\cdot98\cdot100}{4\cdot6\cdot8\cdot10\cdot\ldots\cdot100\cdot102}=\frac{2}{102}=\frac{1}{51}\)
