Đặt C= 1.2+2.3+3.4+...+n.(n+1)
3C=1.2.3+2.3.3+3.4.3+...+n.(n+1).3
3C=1.2.3+2.3.(4-1)+3.4.(5-2)+....+n.(n+1)+[(n+2)-(n-1)]
3C=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)
3C=n.(n+1).(n+2)
C=\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)