$Use induction to prove that for any natural n the following holds:  1\bullet2+2\bullet 3+...+(n-1)\bullet n=\frac{(n-1)n(n+1)}{3} $

We want to prove 
\[1\times 2+2 \times 3+ \dots (n-1)\times n=\frac{(n-1)n(n+1)}{3}.        (*)\]
For $n=2$ (*) is 
\[1\times 2=\frac{(2-1)2(2+1)}{3}=2,\]
which clearly holds. 

Now suppose (*) holds for $n$, we show that it holds for $n+1$. We have

\[1\times 2+2 \times 3+ \dots (n-1)\times n + n \times (n+1)\]
\[=\frac{(n-1)n(n+1)}{3}+ n \times (n+1)\]
\[=\frac{(n-1)n(n+1)+3 n(n+1)}{3}\]
\[=\frac{n(n+1) [(n-1)+3]}{3}=\frac{n(n+1) (n+2)}{3}.\]
Hence we have shown
\[1\times 2+2 \times 3+ \dots (n-1)\times n + n \times (n+1)=\frac{n(n+1) (n+2)}{3}\]
i.e. $(*)$ holds for $n+1$. By induction $(*)$ holds for all $n\geq 2$.