\(\frac{3n}{n+1}=\frac{3n+3-3}{n+1}=\frac{3n+3}{n+1}-\)\(\frac{3}{n+1}=3-\frac{3}{n+1}\)
Để\(\frac{3n}{n+1}\in N\Rightarrow3-\frac{3}{n+1}\in N\Rightarrow\frac{3}{n+1}\in N;\frac{3}{n+1}\le3\)
\(\Rightarrow n+1=1\)hoặc \(n+1=3\)
TH1: \(n+1=1\Rightarrow n=0\)Khi đó: \(\frac{3n}{n+1}=\frac{3.0}{0+1}=0\)
TH2: \(n+1=3\Rightarrow n=2\) Khi đó: \(\frac{3n}{n+1}=\frac{3.2}{2+1}=\frac{6}{3}=2\)