Question

1/1+2+3.1/1+2+3+4+.....+1/1+2+3+4+...+2021

Answer 1

Sửa đề: \(\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+2021}\)

\(=\dfrac{1}{3\cdot\dfrac{4}{2}}+\dfrac{1}{4\cdot\dfrac{5}{2}}+...+\dfrac{1}{2021\cdot\dfrac{2022}{2}}\)

\(=\dfrac{2}{3\times4}+\dfrac{2}{4\times5}+...+\dfrac{2}{2021\times2022}\)

\(=2\left(\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{2021\times2022}\right)\)

\(=2\times\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)

\(=2\times\left(\dfrac{1}{3}-\dfrac{1}{2022}\right)=2\times\dfrac{673}{2022}=\dfrac{673}{1011}\)