$A=1.2+2.3+3.4+…+n.(n+1)$
$⇒3A=3.[1.2+2.3+3.4+…+n.(n+1)]$
$⇒3A=1.2.3+2.3.3+3.4.3+…+n.(n+1).3$
$⇒3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+n.(n+1).[(n+2)-(n-1)]$
$⇒3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n.(n+1).(n+2)-(n-1).n.(n+1)$
$⇒3A=n.(n+1).(n+2)$
$⇒A=\dfrac{n.(n+1).(n+2)}{3}$