Giải thích các bước giải:
`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=(4−0).1.2.3+(5−1).2.3.4+.........+[(n+2)−(n−2)](n−1)n(n+1)`
`⇔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+3)−(n−2)(n−1)n(n+1)`
`⇔4B=(n−1)n(n+1)(n+2)`
`⇔B=` `(n−1)n(n+1)(n+2)/4`