For every n ∊ N, 1・2 + 2・3 + 3・4 + ⋅ ⋅ ⋅ + n(n+1) = n(n+1)(n+2)/3

 Proof by Induction. 

The base case \((n=1):\) \(1*2=1(1+1)(1+2)/3.\) Which simplifies to \(2=2.\)

The induction step. 

The induction hypothesis: assume that for some \(n ∊ N.\) Then \(1*2 + 2*3 + 3*4 +⋅⋅⋅+ n(n+1)=n(n+1)(n+2)/3.\) We must show \(1・2+2・3+3・4+⋅⋅⋅+n(n+1)+(n+1)(n+2)=(n+1)(n+2)(n+3)/3.\) However, \(1*2 + 2*3 + 3*4 +⋅⋅⋅+ n(n+1)=n(n+1)(n+2)/3+(n+1)(n+2).\) \(n(n+1)(n+2)/3+(n+1)(n+2)\) can be rewritten as \(n/3(n+1)(n+2)+(n+1)(n+2).\) Then it can be factored out as \((n+1)(n+2))(n/3+1).\) Finally, it can be rewritten as \((n+1)(n+2)(n+3)/3.\) So, for every \(n ∊ N,\) \(1・2+2・3+3・4+⋅⋅⋅+n(n+1)=(n)(n+1)(n+2)/3.\)