Let S denote the sum
[tex]1\cdot3+2\cdot4+3\cdot5+\cdots+n\cdot(n+2)[/tex]
We can condense this to sigma notation:
[tex]S=\displaystyle\sum_{k=1}^nk(k+2)[/tex]
Expand the summand:
[tex]S=\displaystyle\sum_{k=1}^nk^2+2\sum_{k=1}^nk[/tex]
Recall the Faulhaber formulas,
[tex]\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2[/tex]
[tex]\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6[/tex]
So we have
[tex]S=\dfrac{n(n+1)(2n+1)}6+n(n+1)=\boxed{\dfrac{n(n+1)(2n+7)}6}[/tex]
