Tính H = ( 1 - 1/ 1 + 2 ) x ( 1 - 1/ 1 + 2 + 3 ) x (1 - 1/1 + 2 + 3 + 4 ) x...x ( 1/1 + 2 + ... +2018)
$ H = \left(1 - \dfrac{1}{1 + 2} \right) \left(1 - \dfrac{1}{1 + 2 + 3} \right) \left(1 - \dfrac{1}{1 + 2 + 3 + 4} \right) ... \left(1 - \dfrac{1}{1 + 2 + ... + 2018} \right) \\ = \left[1 - \dfrac{2}{2(1 + 2)} \right] \left[1 - \dfrac{2}{2(1 + 2 + 3)} \right] \left[1 - \dfrac{2}{2(1 + 2 + 3 + 4)} \right]... \left[1 - \dfrac{2}{2(1 + 2 + ... + 2018)} \right] \\ = \left(1 - \dfrac{2}{2 . 3} \right) \left(1 - \dfrac{2}{3 . 4} \right) \left(1 - \dfrac{2}{4 . 5} \right) ... \left(1 - \dfrac{2}{2018 . 2019} \right) \\ = \dfrac{2 . 3 - 2}{2 . 3} . \dfrac{3 . 4 - 2}{3 . 4} . \ ... \ . \dfrac{2018 . 2019 - 2}{2018 . 2019} \\ = \dfrac{1 . 4}{2 . 3} . \dfrac{2 . 5}{3 . 4} . \ ... \ . \dfrac{2017 . 2020}{2018 . 2019} \\ = \dfrac{(1 . 2 . \ ... \ . 2017)(4 . 5 . \ ... \ . 2020)}{(2 . 3 . \ ... \ . 2018)(3 . 4 . \ ... \ . 2019)} \\ = \dfrac{1 . 2020}{2018 . 3} \\ = \dfrac{1010}{3027} $
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
