Question

[oOo FC Beerus sama oOo]

Cho \(M=2009-\frac{1}{8}-\frac{2}{9}-\frac{3}{10}-\frac{4}{11}-..........-\frac{2009}{2016}\) và \(N=\frac{1}{24}+\frac{1}{27}+\frac{1}{30}+............+\frac{1}{6048}\)

Tính \(\frac{M}{N}\)

Answer 1

[evermore Mathematics]

ta có : M = \(\left(1-\frac{1}{8}\right)+\left(1-\frac{2}{9}\right)+\left(1-\frac{3}{10}\right)+...+\left(1-\frac{2009}{2016}\right)\)

M = \(\frac{7}{8}+\frac{7}{9}+\frac{7}{10}+...+\frac{7}{2016}\)

M = \(\frac{21}{24}+\frac{21}{27}+\frac{21}{30}+........+\frac{21}{6048}\)

M = \(21\cdot\left(\frac{1}{24}+\frac{1}{27}+\frac{1}{30}+....+\frac{1}{6048}\right)\)

=> \(\frac{M}{N}=21\)