Question

[Ngô Ngọc Vinh]

Tìm X ,biêt  1+1/3 +1/6+1/10+...+2/X.(x+1) =1  2025/2027

Answer 1

[Nguyễn Lê Phước Thịnh]

\(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{2025}{2027}\)

=>\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4052}{2027}\)

=>\(2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{4052}{2027}\)

=>\(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2026}{2027}\)

=>\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2026}{2027}\)

=>\(1-\dfrac{1}{x+1}=\dfrac{2026}{2027}\)

=>\(\dfrac{1}{x+1}=\dfrac{1}{2027}\)

=>x+1=2027

=>x=2026