[tex]1\cdot 3+3\cdot 5+...+(2n-1)\cdot (2n+1) = \dfrac{n\cdot (4n^2+6n-1)}{3}\\ \\\\\displaystyle \sum\limits_{k=1}^{n}\Big[(2k-1)(2k+1)\Big] =\sum\limits_{k=1}^{n}(4k^2-1) =\\ \\ =4\sum\limits_{k=1}^n k^2-\sum\limits_{k=1}^n1 = 4\cdot \dfrac{n(n+1)(2n+1)}{6}-n= \\ \\ = 2\cdot \dfrac{n(n+1)(2n+1)}{3}-n =2\cdot \dfrac{n(n+1)(2n+1)-3n}{3}= \\ \\ =\dfrac{n\Big[2(n+1)(2n+1)-3\Big]}{3} = \dfrac{n\Big[2(2n^2+3n+1)-3\Big]}{3}=\\ \\ = \boxed{\dfrac{n(4n^2+6n-1)}{3}}[/tex]
