Question

\(ChoM=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{999}\\ N=\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\\ Tính\frac{M}{N}\)

Answer 1

\(N=\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\)

\(1000N=1+\frac{1}{999}+\frac{1}{3}+\frac{1}{997}+...+\frac{1}{997}+\frac{1}{3}+1\)

\(1000N=2\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)

\(N=\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)

\(\frac{M}{N}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]}=\frac{1}{\frac{1}{50}}=50\)