Đáp án: $A=\dfrac{n+2}{3n}$
Giải thích các bước giải:
Ta có :
$\begin{split}1-\dfrac{1}{1+2+3+..+n}&=1-\dfrac{1}{\dfrac{n(n+1)}{2}}\\&=1-\dfrac{2}{n(n+1)}\\&=\dfrac{n(n+1)-2}{n(n+1)}\\&=\dfrac{n^2+n-2}{n(n+1)}\\&=\dfrac{(n-1)(n+2)}{n(n+1)}\end{split}$
$\begin{split}\rightarrow A&= (1-\dfrac{1}{1+2})(1-\dfrac{1}{1+2+3})..(1-\dfrac{1}{1+2+3+..+2020})\\&=\dfrac{(2-1)(2+2)}{2.(2+1)}.\dfrac{(3-1)(3+2)}{3.(3+1)}.\dfrac{(4-1)(4+2)}{4.(4+1)}..\dfrac{(n-1)(n+2)}{n(n+1)}\\ &=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}..\dfrac{(n-1)(n+2)}{n(n+1)}\\ &=\dfrac{(1.2.3...(n-1)).(4.5.6...(n+2))}{(2.3.4...n).(3.4.5..(n+1))}\\ &=\dfrac{n+2}{3n}\end{split}$
