Question

tìm số tự nhiên x biết 1/3+1/6+1/10+...+2/x.(x+1)=2020/2022

Answer 1

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{505}{1011}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1010}{1011}\)

=>1/x+1=-1009/2022

=>x+1=-2022/1009

hay x=-3031/1009