Question

[Yến Nguyễn]

Thực hiện phép tính

( 1 - \(\frac{1}{1+2}\)) . ( 1 - \(\frac{1}{1+2+3}\)) ... ( 1 - \(\frac{1}{1+2+3+...+2018}\))

Answer 1

[Stephen Hawking]

\(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).....\left(1-\frac{1}{1+2+3+.....+2018}\right)\)

\(=\left(1-\frac{1}{\frac{2.3}{2}}\right).\left(1-\frac{1}{\frac{3.4}{2}}\right).......\left(1-\frac{1}{\frac{2018.2019}{2}}\right)\)

\(=\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).......\left(1-\frac{2}{2018.2019}\right)\)

\(=\left(1-\frac{1}{3}\right).\left(1-\frac{5}{6}\right).......\left(1-\frac{1}{2037171}\right)\)

\(=\frac{2}{3}.\frac{5}{6}......\frac{2037170}{2037171}\)

\(=\frac{4}{6}.\frac{10}{12}.......\frac{4074340}{4074342}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}......\frac{2017.2020}{2018.2019}\)

\(=\frac{1.2......2017}{2.3.....2018}.\frac{4.5......2020}{3.4......2019}=\frac{1}{2018}.\frac{2020}{3}=\frac{1010}{3027}\)