Question

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

Answer 1

\(B=1.2.3+2.3.4+...+\left(n-1\right)n\left(n+1\right)\\ \Rightarrow4B=1.2.3.4+2.3.4.4+...+\left(n-1\right)n\left(n+1\right).4\\ \Rightarrow4B=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+...+\left(n-1\right)n\left(n+1\right).\left[\left(n+2\right)-\left(n-2\right)\right]\)

\(\Rightarrow4B=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+\left(n-1\right)n\left(n+1\right).\left(n+2\right)-\left(n-2\right)\left(n-1\right)n\left(n+1\right)\)

\(\Rightarrow4B=\left(n-1\right)n\left(n+1\right).\left(n+2\right)\\ \Rightarrow B=\dfrac{\left(n-1\right)n\left(n+1\right).\left(n+2\right)}{4}\)

Answer 2

\(B=1\times2\times3+2\times3\times4+...+\left(n-1\right)n\left(n+1\right)\)

\(\Rightarrow4B=1\times2\times3\times4+2\times3\times4\times4+...+\left(n-1\right)n\left(n+1\right).4\)

\(=1\times2\times3\times4-0\times1\times2\times3+2\times3\times4\times5-1\times2\times3\times4+...+\left(n-1\right)n\left(n+1\right)\times\left(n+2\right)-\left[\left(n-2\right)\times\left(n-1\right)n\left(n+1\right)\right]\)\(=\left(n-1\right)n\left(n+1\right)\times\left(n+2\right)-0\times1\times2\times3=\left(n-1\right)n\left(n+1\right)\times\left(n+2\right)\)

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