[tex]1\cdot2+2\cdot3+\cdots+(n-1)\cdot n+n\cdot(n+1)=\displaystyle\sum_{k=1}^nk(k+1)=\sum_{k=1}^n(k^2+k)[/tex]
[tex]=\dfrac{n(n+1)(2n+1)}6+\dfrac{n(n+1)}2[/tex]
[tex]=\dfrac{n(n+1)(2n+1)+3n(n+1)}6[/tex]
[tex]=\dfrac{2n^3+6n^2+4n}6[/tex]
[tex]=\dfrac13n^3+n^2+\dfrac23n[/tex]
