Question

Tính

\(\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)...\left(1-\dfrac{1}{1+2+3+...+2016}\right)\)

Answer 1

\(\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)...\cdot\left(1-\dfrac{1}{1+2+3+...+2016}\right)\)

= \(\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)...\left(1-\dfrac{2}{2016.2017}\right)\)

= \(\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}...\dfrac{2014.2017}{2015.2016}.\dfrac{2015.2018}{2016.2017}\)

= \(\dfrac{2018}{3.2016}\)

= \(\dfrac{1009}{3024}\)

Answer 2

\(\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)...........\left(1-\dfrac{1}{1+2+3+.........+2016}\right)\)

\(=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right).............\left(1-\dfrac{1}{2033136}\right)\)

\(=\dfrac{2}{3}.\dfrac{5}{6}.............\dfrac{2033135}{2033136}\)

Sao nữa? =))