Giả sử \(ƯCLN\left(n,2n+1\right)=d\)
\(\Rightarrow\hept{\begin{cases}n⋮d\\2n+1⋮d\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2n⋮d\\2n+1⋮d\end{cases}}\)
\(\Rightarrow2n+1-2n⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
\(\RightarrowƯCLN\left(2n+1,n\right)=1\)
Vậy \(ƯCLN\left(2n+1,n\right)=1\)với mọi \(n\in N\)