Question

Cho:P =(1-1/1+2)(1-1/1+2+3)(1-1/1+2+3+4)...(1-1/1+2+3+...+2014). Tính: 2014/2016P

Answer 1

\(P=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+2014}\right)\)

\(P=\frac{\left(1+2\right).2:2-1}{\left(1+2\right).2:2}.\frac{\left(1+3\right).3:2-1}{\left(1+3\right).3:2}.\frac{\left(1+4\right).4:2-1}{\left(1+4\right).4:2}...\frac{\left(1+2014\right).2014:2-1}{\left(1+2014\right).2014:2}\)

\(P=\frac{2}{2.3:2}.\frac{5}{3.4:2}.\frac{9}{4.5:2}...\frac{2029104}{2014.2015:2}\)

\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{2013.2016}{2014.2015}\)

\(P=\frac{1.2.3...2013}{2.3.4...2014}.\frac{4.5.6...2016}{3.4.5...2015}\)

\(P=\frac{1}{2014}.\frac{2016}{3}=\frac{1}{2014}.672=\frac{336}{1007}\)