Cách giải:
$A=1.2.3+2.3.4+.......+n(n+1)(n+2)$
$\to 4A=1.2.3.4+2.3.4.4+.....+n(n+1)(n+2).4$
$\to 4A=1.2.3.4+2.3.4.(5-1)+.....+n(n+1)(n+2).[n+3-(n-1)]$
$\to 4A=1.2.3.4+2.3.4.5-1.2.3.4+.....+n(n+1)(n+2)(n+3)-(n-1).n(n+1)(n+2)$
$\to 4A=n(n+1)(n+2)(n+3)$
$\to 4A+1=[(n+1)(n+2)][n(n+3)]+1$
$\to 4A+1=(n^2+3n+2)(n^2+3n)+1$
$\to 4A+1=(n^2+3n+1+1)(n^2+3n+1-1)+1$
$\to 4A+1=(n^2+3n+1)^2-1+1$
$\to 4A+1=(n^2+3n+1)^2$ là 1 số chính phương.
