$\text{Câu 18:}$
$\dfrac{1}{1+2}$+$\dfrac{1}{1+2+3}$+....+ $\dfrac{1}{1+...+2020}$
⇒$\dfrac{1}{2·(1+2)}$+$\dfrac{1}{3·(1+3}$+....+$\dfrac{1}{2020·(2020+1)}$ ($\text{trong ngoặc tròn bạn lấy số đầu+số cuối}$)
⇒$\dfrac{1}{2·3}$+ $\dfrac{1}{3·4}$+....+$\dfrac{1}{2020·2021}$
$ADCT$:$\dfrac{k}{n·(n+k)}$=$\dfrac{1}{n}$ -$\dfrac{1}{n+k}$
⇒ $\dfrac{1}{2}$-$\dfrac{1}{3}$+$\dfrac{1}{3}$+...+$\dfrac{-1}{2021}$
⇒ $\dfrac{1}{2}$+$0$+...+ $\dfrac{-1}{2021}$
⇒$\dfrac{1}{2}$- $\dfrac{1}{2021}$
⇒$\dfrac{2019}{4042}$
