Ta có :
B = 1 . 2 . 3 + 2 . 3 . 4 + . . . + ( n - 1 ) n ( n + 1 )
⇒ 4B = 1 . 2 . 3 . 4 + 2 . 3 . 4 . 4 + . . . + ( n - 1 ) n ( n + 1 ) . 4
4B = 1 . 2 . 3 . 4 - 0 . 1 . 2 . 3 + 2 . 3 . 4 . 5 - 1 . 2 . 3 . 4 + . . . + ( n - 1 ) n ( n + 1 ) ( n + 2 ) - [ ( n - 2 ) ( n - 1 ) n ( n + 1 ) ]
4B = ( n - 1 ) n ( n + 1 ) ( n + 2 ) - 0 . 1 . 2 . 3 = ( n - 1 ) n ( n + 1 ) ( n + 2 )
⇒ B = `[(n-1)n(n+1)(n+2)]/4`
Vậy B = `[(n-1)n(n+1)(n+2)]/4`
