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.

Ta có: A=1.2.3 + 2.3.4 + ... + n(n+1)(n+2)

⇔4A=1.2.3.4 + 2.3.4.4 + ... + n(n+1)(n+2).4

⇔4A=1.2.3.4 + 2.3.4.(5-1) + ... + n(n+1)(n+2)[(n+3)-(n-1)]

⇔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)

⇔4A= n(n+1)(n+2)(n+3)

⇔4A=(n²+3n)(n²+3n+2)

⇔4A= (n²+3n)² + 2(n²+3n)

⇔4A+1=(n²+3n)² + 2(n²+3n) +1

⇔4A+1=(n²+3n+1)² là số chính phương. ⇒ đpcm