Argentine National Olympiad 2015

Denote $S_n={\sum }_{k=1}^n\frac{k(3k+1)}{(3k-1)(3k+2)}$. Multiply all numerators by $12$ and all denominators by $4$ to obtain $$3S_n=\sum _{k=1}^n\frac{12k(3k+1)}{(6k-2)(6k+4)}.$$ Now complete squares as follows: $$\begin{array}{rl}12k(3k+1)& =36k^2+12k=(6k+1{)}^2-1,\\ (6k-2)(6k+4)& =36k^2+12k-8=(6k+1{)}^2-9.\end{array}$$ Therefore $$\frac{12k(3k+1)}{(6k-2)(6k+4)}=\frac{(6k+1{)}^2-1}{(6k+1{)}^2-9}=1+\frac{8}{(6k+1{)}^2-9}=1+\frac{8}{(6k-2)(6k+4)}$$ And it follows that $3S_n=n+8{\sum }_{k=1}^n\frac{1}{(6k-2)(6k+4)}$. Since $$\frac{1}{(6k-2)(6k+4)}=\frac{1}{6}\left(\frac{1}{6k-2}-\frac{1}{6k+4}\right),$$ we obtain $$3S_n=n+8\cdot \frac{1}{6}\left(\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{16}+\cdots +\frac{1}{6n-2}-\frac{1}{6n+4}\right)=n+\frac{4}{3}\left(\frac{1}{4}-\frac{1}{6n+4}\right)$$ This leads to $3S_n=n+\frac{n}{3n+2}$. For $n=99$ the answer is $$S_{99}=33+\frac{33}{299}=\frac{9900}{299}.$$