Sum square difference

The sum of the n first natural numbers is: $R_{n}=\frac{n(n+1)}{2}$ . The sum of the squares of the n first natural numbers is: $S_{n}=\frac{n(n+1)(2n +1)}{6}$ We want to evaluate the quantity $(R_{n})^{2}-S_{n}$ for n=100. $(R_{n})^{2}-S_{n}=$ $(\frac{n(n+1)}{2})^{2}-\frac{n(n+1)(2n +1)}{6}=$ $\frac{n^{2}(n+1)^{2}}{4}-\frac{n(n+1)(2n +1)}{6}=$ $\frac{n(n+1)}{2}(\frac{n(n+1)}{2}-\frac{2n+1}{3})=$ $\frac{n(n+1)}{2}\frac{3n^{2}+3n-4n-2}{6}=$ $\frac{n(n+1)(3n^{2}-n-2)}{12}=$ $\frac{3n^{4}+2n^{3}-3n^{2}-2n}{12}$ . For n=100, this works out to be 25164150.