B = 1.2.3 + 2.3.4 + ... + (n - 1)n(n + 1)

We have:

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)

=>B = \(\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)

Ta có :           

\(B=1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\)

\(4B=1.2.3.4+2.3.4.4+...+\left(n-1\right).n.\left(n+1\right).4\)

\(4B=1.2.3.4+2.3.4\left(5-1\right)+...+\left(n-1\right).n.\left(n+1\right).\left[\left(n+2\right)-\left(n-2\right)\right]\)

\(4B=1.2.3.4+2.3.4.5-1.2.3.4+...+n.\left(n+1\right).\left(n+2\right)-\left(n-1\right).n.\left(n+1\right)\)

\(4B=\left[1.2.3.4+2.3.4.5+...+n.\left(n+1\right).\left(n+2\right)\right]\)\(-\left[1.2.3.4+2.3.4.5+...+\left(n-1\right).n.\left(n+1\right)\right]\)

\(4B=n.\left(n+1\right).\left(n+2\right)\)

\(B=\dfrac{n.\left(n+1\right).\left(n+2\right)}{4}\)