Question

Giúp mik!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

\(P=\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+4+...+2021}\\ P=1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{2021\cdot2022:2}\\ \dfrac{P}{2}=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2021\cdot2022}\\ \dfrac{P}{2}=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2021\cdot2022}\\ \dfrac{P}{2}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\\ \dfrac{P}{2}=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\\ P=\dfrac{2021}{2022}:2=\dfrac{2021}{1011}\)