Question

Aratati ca n(n+1)(n+2)(n+3)+1=(n•n+3n+1)(n•n+3n+1)

Answer 1

$n(n+1)(n+2)(n+3)+1= \\ = (n^2+n)(n+2)(n+3)+1 \\ =(n^3+2n^2+n^2+2n)(n+3)+1 \\ =(n^3+3n^2+2n)(n+3)+1 \\ =n^4+3n^3+2n^2+3n^3+9n^2+6n+1 \\ =n^4+9n^2+1+6n^3+6n+2n^2 \\ =(n^2+3n+1)^2$