Ta có, với \(n\) nguyên dương: \(1+2+...+n=\frac{n\left(n+1\right)}{2}\)
Suy ra, \(1-\frac{1}{1+2+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Khi đó:
\(1-\frac{1}{1+2}=\frac{1.4}{2.3}\)
\(1-\frac{1}{1+2+3}=\frac{2.5}{3.4}\)
....
\(1-\frac{1}{1+2+...+2013}=\frac{2012.2015}{2013.2014}\)
\(1-\frac{1}{1+2+...+2014}=\frac{2013.2016}{2014.2015}\)
Suy ra, \(P=\frac{\left(1.2.....2013\right).\left(4.5.....2016\right)}{2.\left(3.4.....2014\right)^2.2015}=\frac{2016}{3.2014}=\frac{336}{1007}\)
Bạn k làm thì đừng trả lời linh tinh nhá !