Sum 'em Up 2

We can rewrite the summation as $\begin{aligned} \displaystyle\sum_{r=1}^{2016}\frac{r(r+1)(2r+1)/6}{r^2(r+1)^2/4}&=\frac 43\eta-\frac k{6051}\\ \frac 23\displaystyle\sum_{r=1}^{2016}\frac{(2r+1)}{(r+1)r}&=\frac 23\displaystyle\sum_{r=1}^{2016}\frac 2r-\frac k{6051}\\ \frac 23\displaystyle\sum_{r=1}^{2016}\frac{(r+1)-r}{(r+1)r}-\frac 2r&=\frac k{6051}\\ \frac 23\displaystyle\sum_{r=1}^{2016}\frac 1r-\frac 1{(r+1)}&=\frac k{6051}\\ \frac 23\left(1-\frac 1{2017}\right)&=\frac k{6051}\\ k&=4032\\ \end{aligned}$

The general term of the series can be written as $\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n^2(n+1)^2}{4}}$ . Therefore, $\sum^{2016}_{n=1} \frac{2(2n+1)}{3n(n+1)}$ $=\frac{2}{3} \sum^{2016}_{n=1} \frac{2n+1}{n(n+1)}$ $=\frac{2}{3} \sum^{2016}_{n=1} \frac{(n+1)+n}{n(n+1)}$ $=\frac{2}{3} \sum^{2016}_{n=1} \frac{1}{n}+\frac{1}{n+1}$ $=\frac{2}{3}(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017})$ $=\frac{2}{3}(1+2(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016})+\frac{1}{2017})$ $=\frac{2}{3}(1+2(\eta-1)+\frac{1}{2017})$ $=\frac{2}{3}(1+2\eta-2+\frac{1}{2017})$ $=\frac{2}{3}(2\eta-\frac{2016}{2017})$ $=\frac{4\eta}{3}-\frac{4032}{6051}$ Therefore, $\boxed{k=4032}$