Tính C = 1.2 + 2.3 + 3.4 +….+ n(n + 1)

Ta có:

3C = 1.2.3 + 2.3.3 + 3.4.3 + … + n.(n + 1).3

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 = \[\frac{{n(n + 1)(n + 2)}}{3}\].