Resposta :
- [tex]n!=n.(n-1)(n-2)...(n-k)...1[/tex]
- [tex]a.b+ac=a(b+c)[/tex]
[tex]\frac{(k+2)!-k!}{(k+1)!-k!}\\\\\frac{(k+2)(k+1)k!-k!}{(k+1)k!-k!}\\\\\frac{k![(k+2)(k+1)-1]}{k![(k+1)-1]}\\\\\frac{(k+2)(k+1)-1}{(k+1)-1}\\\\\frac{k^2+k+2k+2-1}{k}\\\\\frac{k^2+3x+1}{k}[/tex]