Bonjour,
[tex] \dfrac{n^2-1}{n^2} =\dfrac{n+1}{2n} \\\\\Rightarrow\ (n^2-1)*2n=(n+1)*n^2\\\\\Rightarrow\ 2n^3-2n=n^3+n^2\\\\\Rightarrow\ n^3-n^2-2n=0\\\\\Rightarrow\ n(n^2-n-2)=0\\\\\Rightarrow\ n(n^2-2n+n-2)=0\\\\\Rightarrow\ n(n-2)(n+1)=0\\\\\Rightarrow\ n=-1\ ou\ n=0\ ou\ n=2\\\\ [/tex]