Câu hỏi của nguyễn hà khởi my

Ta có:\(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\Rightarrow\dfrac{1}{1+2+3+...+n}=\dfrac{2}{n\left(n+1\right)}=\dfrac{2}{n}-\dfrac{2}{n+1}\)Do đó:\(P=\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+...+\dfrac{2}{2021}-\dfrac{2}{2022}\)\(P=\dfrac{2}{2}-\dfrac{2}{2022}=1-\dfrac{1}{1011}=\dfrac{1010}{1011}\)Câu trả lời: Ta có:\(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\Rightarrow\dfrac{1}{1+2+3+...+n}=\dfrac{2}{n\left(n+1\right)}=\dfrac{2}{n}-\dfrac{2}{n+1}\)Do đó:\(P=\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+...+\dfrac{2}{2021}-\dfrac{2}{2022}\)\(P=\dfrac{2}{2}-\dfrac{2}{2022}=1-\dfrac{1}{1011}=\dfrac{1010}{1011}\)