Question

(fei-sp) simplificando (n+2)! + (n+1)!/ (n+3)!, (n€N) obtém-se?

Answer 1

Imagino que deve ser assim

[(n+2)!+(n+1)!]/(n+3)!=[ (n+2)(n+1)!+(n+1)! ]/(n+3)!

= (n+1)!(n+2+1)/(n+3)!=(n+1)!(n+3)/(n+3)!=(n+1)!(n+3)/(n+3)(n+2)(n+1)!

simplificando temos

=1/(n+2)