Question

Tính B=\(\left(1-\dfrac{1}{1+2}\right).\left(1-\dfrac{1}{1+2+3}\right).....\left(1-\dfrac{1}{1+2+3+...+2017}\right)\)

Answer 1

Ta có \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\Rightarrow\dfrac{1}{1+2+...+n}=\dfrac{2}{n\left(n+1\right)}\)

\(\Rightarrow1-\dfrac{1}{1+2+...+n}=1-\dfrac{2}{n\left(n+1\right)}=\dfrac{n^2+n-2}{n\left(n+1\right)}=\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

Áp dụng vào bài toán:

\(B=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{2015.2018}{2016.2017}.\dfrac{2016.2019}{2017.2018}\)

\(B=\dfrac{1.2.3...2016}{2.3.4...2017}.\dfrac{4.5.6...2019}{3.4.5...2018}=\dfrac{1}{2017}.\dfrac{2019}{3}=\dfrac{673}{2017}\)