Giải thích các bước giải:
Bài 2:
a.Ta có :
$\begin{split}1-\dfrac{1}{1+2+3+..+n}&=1-\dfrac{1}{\dfrac{n(n+1)}{2}}\\&=1-\dfrac{2}{n(n+1)}\\&=\dfrac{n(n+1)-2}{n(n+1)}\\&=\dfrac{n^2+n-2}{n(n+1)}\\&=\dfrac{(n-1)(n+2)}{n(n+1)}\end{split}$
$\rightarrow A=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}....\dfrac{2019.2022}{2020.2021}$
$\rightarrow A=\dfrac{(1.2.3.. .2019).(4.5.6...2022)}{(2.3..2020).(3.4...2021)}$
$\rightarrow A=\dfrac{2022}{2020.3}$
$\rightarrow A=\dfrac{337}{1010}$