Question

tính A biết

A = \(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+......+2018}\right)\)

ai làm đúng cho 2 tick luôn

thanks

Answer 1

bằng 1/123

Answer 2

\(Taco\):

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

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

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

\(\Rightarrow A+2017.\left(\frac{1}{3}\right).....\frac{2+3+.....+2018}{1+2+3+...+2018}=1.1.1......1=1\)

\(.................................\)

Answer 3

\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right).....\left(1-\frac{1}{1+2+...+2018}\right)\)

\(A=\left(1-\frac{1}{\frac{2\left(2+1\right)}{2}}\right)\left(1-\frac{1}{\frac{3\left(3+1\right)}{2}}\right)\left(1-\frac{1}{\frac{4\left(4+1\right)}{2}}\right).....\left(1-\frac{1}{\frac{2018\left(2018+1\right)}{2}}\right)\)

\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right).....\left(1-\frac{1}{2037171}\right)\)

\(A=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.....\frac{2037170}{2037171}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.....\frac{4074340}{4074342}\)

\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{2017.2020}{2018.2019}=\frac{1.2.3.4.....2017}{2.3.4.5.....2018}.\frac{4.5.6.....2020}{3.4.5.....2019}\)

\(A=\frac{1}{2018}.\frac{2020}{3}=\frac{1010}{3027}\)

PS : ko chắc :v có làm vài lần nhưng quên ko nhớ rõ cách giải