\(\frac{3n+1}{2n+1}=\frac{n+n+n+1}{n+n+1}=\frac{2n+n+1}{2n+1}=\frac{2n+1}{2n+1}+\frac{1}{2n+1}=1+\frac{1}{2n+1}\)
Để \(\frac{1}{2n+1}\) nguyên thì \(2n+1\in\text{Ư}\left\{1;-1\right\}\)
\(2n+1=1\Rightarrow2n=0\Rightarrow n=0\)\(2n+1=-1\Rightarrow2n=-2\Rightarrow n=-1\)Vậy: \(n\in\left\{-1;0\right\}\)