Pentru ipoteza (nescrisă) că n ∈ ℕ*, vom avea:
[tex]\it \dfrac{3n+1}{n}=\dfrac{3n}{n}+\dfrac{1}{n}=3+\dfrac{1}{n}=3\dfrac{1}{n}[/tex]
[tex]\it \dfrac{5n+7}{n+1}=\dfrac{5n+5+2}{n+1}= \dfrac{5n+5}{n+1}+\dfrac{2}{n+1}=\dfrac{5(n+1)}{n+1}+\dfrac{2}{n+1}=5\dfrac{2}{n+1}[/tex]
