I have come across a problem in my homework that describes the sum of a binomial squared, and I can't think of a way to simplify it. I have an idea that it would involve $\frac{\left(n\right)\left(n+1\right)\left(2n+1\right)}{6}$, but I can't seem to format the equation in a way that would be helpful. Any help would be much appreciated. There are two problems below, a solution to either would be amazing.
$\sum_{i=0}^{n-1}\frac{\left(\frac{\left(n+1\right)}{2}-i\right)^2}{n}$ $\sum_{i=0}^{n-1}\left(\frac{\left(n+1\right)}{2}-i\right)^2$
$\begin{array}\\ s(n) &=\sum_{i=0}^{n-1}\left(\frac{\left(n+1\right)}{2}-i\right)^2\\ &=\frac14\sum_{i=0}^{n-1}\left(n+1-2i\right)^2\\ &=\frac14\sum_{i=0}^{n-1}((n+1)^2-4(n+1)i+4i^2)\\ &=\frac14\sum_{i=0}^{n-1}(n+1)^2-\sum_{i=0}^{n-1}(n+1)i+\sum_{i=0}^{n-1}i^2\\ &=\frac14n(n+1)^2-\frac12 n(n-1)(n+1)+\frac16 (n-1)n(2n-1)\\ &=\dfrac{n}{12}\left( 3(n+1)^2-6(n-1)(n+1)+2 (n-1)(2n-1)\right)\\ &=\dfrac{n}{12}\left( 3n^2+6n+3-6(n^2-1)+2(2n^2-3n+1)\right)\\ &=\dfrac{n}{12}\left( 3n^2+6n+3-6n^2+6+4n^2-6n+2\right)\\ &=\dfrac{n}{12}\left( n^2+11\right)\\ \end{array} $