\(A=\left(1+\frac{1}{2^2-1}\right)\left(1+\frac{1}{3^2-1}\right)\left(1+\frac{1}{4^2-1}\right)\cdot...\cdot\left(1+\frac{1}{100^2-1}\right)\)
\(=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot...\cdot\frac{99^2}{98\cdot100}\cdot\frac{100^2}{99\cdot101}=\frac{200}{101}\)
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)......\left(1+\frac{1}{99.100}\right)\)
\(=\left(1+\frac{1}{2^2-1}\right)\left(1+\frac{1}{3^2-1}\right)......\left(1+\frac{1}{100^2-1}\right)\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}..............\frac{100^2}{99.100}=\frac{200}{101}\)
T nha
