Đặt \(B=\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\cdots+\frac{1}{99\cdot1}\)
\(=2\left(\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\cdots+\frac{1}{49\cdot51}\right)\)
\(=\frac{2}{100}\left(\frac{100}{1\cdot99}+\frac{100}{3\cdot97}+\cdots+\frac{100}{49\cdot51}\right)\)
\(=\frac{1}{50}\left(1+\frac{1}{99}+\frac13+\frac{1}{97}+\cdots+\frac{1}{49}+\frac{1}{51}\right)=\frac{1}{50}\left(1+\frac13+\cdots+\frac{1}{97}+\frac{1}{99}\right)\)
Ta có: \(A=\frac{1+\frac13+\frac15+\cdots+\frac{1}{99}}{\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\cdots+\frac{1}{99\cdot1}}\)
\(=\frac{1+\frac13+\cdots+\frac{1}{99}}{\frac{1}{50}\left(1+\frac13+\cdots+\frac{1}{99}\right)}\)
\(=1:\frac{1}{50}=50\)
