Question

Tính :

P = \(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+2017}\right)\)

Answer 1

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

\(P=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)...\left(1-\frac{2}{2017.2018}\right)\)

\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{2016.2019}{2017.2018}\)

\(P=\frac{1.2.3....2016}{2.3.4...2017}.\frac{4.5.6...2019}{3.4.5...2018}=\frac{2019}{3.2017}=\frac{673}{2017}\)