=\(\frac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)
4B = 1.2.3.4 + 2.3.4.4 + ... + (n-1)n(n+1).4
= 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)]
= (n-1)n(n+1)(n+2) - 0.1.2.3
= (n-1)n(n+1)(n+2)
suy ra \(B = {(n-1)n(n+1)(n+2)\over 4}\)