Ta có : $A=1.2.3 +2.3.4+...+k.(k+1).(k+2)$
$⇒4A = 1.2.3.4 + 2.3.4.4 +...+k.(k+1).(k+2).4 $
$=1.2.3.(4-0) + 2.3.4.(5-1) + .... + k.(k+1).(k+2).[(k+3)-(k-1)]$
$=1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + k.(k+1).(k+2).(k+3) - (k-1).k.(k+1).(k+2) $
$= 1.2.3.4+2.3.4.5-1.2.3.4+...+k(k+1).(k+2).(k+3)- (k-1).k.(k+1).(k+2) $
$=k.(k+1).(k+2).(k+3)$
$⇒ 4A + 1 = k(k+1).(k+2).(k+3)+1 $
$= [k.(k+3)].[(k+1).(k+2)] + 1 $
$=(k^2+3k).(k^2+3k+2) + 1 $
$ = (k^2+3k)^2 + 2.(k^2+3k) + 1$
$= (k^2+3k+1)^2$ luôn là một số chính phương (đpcm)
Chúc bạn học tốt !!
