Đặt : `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)`
`⇒A={n(n+1)(n+2).(n+3)}/4` `(đpcm)`