Répondre :
Bonsoir,
[tex]U_n=n^2-3n+1\\\\U_{n-1}=(n-1)^2-3(n-1)+1\\=n^2-2n+1-3n+3+1=n^2-5n+5\\\\U_{n+1}=(n+1)^2-3(n+1)+1\\=n^2+2n+1-3n-3+1=n^2-n-1[/tex]
[tex]U_n=n^2-3n+1\\\\U_{n-1}=(n-1)^2-3(n-1)+1\\=n^2-2n+1-3n+3+1=n^2-5n+5\\\\U_{n+1}=(n+1)^2-3(n+1)+1\\=n^2+2n+1-3n-3+1=n^2-n-1[/tex]