Question

tính B\(=\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right).....\left(1-\dfrac{1}{1+2+...+2018}\right)\)

Answer 1

Ta có:

\(1-\dfrac{1}{1+2+...+n}=1-\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=1-\dfrac{2}{n\left(n+1\right)}=\dfrac{n^2+n-2}{n\left(n+1\right)}=\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

Áp dụng ta được:

\(B=\dfrac{1\cdot4}{2\cdot3}\cdot\dfrac{2\cdot5}{3\cdot4}\cdot...\cdot\dfrac{2017\cdot2020}{2018\cdot2019}=\dfrac{1\cdot2\cdot3\cdot\left(4\cdot5\cdot...\cdot2017\right)^2\cdot2018\cdot2019\cdot2020}{2\cdot\left(3\cdot4\cdot...\cdot2018\right)^2\cdot2019}=\dfrac{1\cdot2\cdot3\cdot2018\cdot2019\cdot2020}{2\cdot3^2\cdot2018^2\cdot2019}=\dfrac{6\cdot2020}{2\cdot3\cdot2018}=\dfrac{2020}{2018}=\dfrac{1010}{1009}\)