Partial fraction challenge (2)

$\begin{aligned} S & = \sum_{n=1}^{2016} \frac n{\color{#3D99F6}{n^4+n^2+1}} \quad \quad \small \color{#3D99F6}{\text{Note that }n^4+n^2+1=(n^2-n+1)(n^2+n+1)} \\ & = \sum_{n=1}^{2016} \frac n{\color{#3D99F6}{(n^2-n+1)(n^2+n+1)}} \\ & = \frac 12 \sum_{n=1}^{2016} \left( \frac 1{n^2-n+1} - \frac 1{n^2+n+1} \right) \\ & = \frac 12 \left( \sum_{n=1}^{2016} \frac 1{n^2-n+1} - \sum_{\color{#3D99F6}{n=1}}^{2016} \frac 1{\color{#3D99F6}{n}^2+\color{#3D99F6}{n}+1} \right) \quad \quad \small \color{#3D99F6}{\text{Let }n=m-1} \\ & = \frac 12 \left( \sum_{n=1}^{2016} \frac 1{n^2-n+1} - \sum_{\color{#3D99F6}{m=2}}^{\color{#3D99F6}{2017}} \frac 1{\color{#3D99F6}{(m-1)}^2+\color{#3D99F6}{(m-1)}+1} \right) \\ & = \frac 12 \left( \sum_{n=1}^{2016} \frac 1{n^2-n+1} - \sum_{m=2}^{2017} \frac 1{m^2-m+1} \right) \quad \quad \small \color{#3D99F6}{\text{Putting }m=n} \\ & = \frac 12 \left( \sum_{n=1}^{2016} \frac 1{n^2-n+1} - \sum_{n=2}^{2017} \frac 1{n^2-n+1} \right) \\ & = \frac 12 \left( 1 - \frac 1{(2017)^2-2017+1} \right) \\ & = \frac 12 \left( 1 - \frac 1{(2017)(2016)+1} \right) \\ & = \frac 12 \left( \frac {(2017)(2016)}{(2017)(2016)+1} \right) \\ & = \frac {(2017)(1008)}{(2017)(2016)+1} \end{aligned}$

$\implies p+q = 1008 + 2016 = \boxed{3024}$

$k^{4}+k^{2}+1=(k^{2}-k+1)(k^{2}+k+1)$

$\frac{k}{k^{4}+k^{2}+1}=\frac{1}{2}(\frac{1}{k^{2}-k+1}-\frac{1}{k^{2}+k+1})$

$S=\displaystyle\sum_{n=1}^{2016}\frac{k}{k^{4}+k^{2}+1}$

$S=\displaystyle\sum_{n=1}^{2016}\frac{1}{2}(\frac{1}{k^{2}-k+1}-\frac{1}{k^{2}+k+1})$

$S=\frac{1}{2}(\displaystyle\sum_{n=1}^{2016}\frac{1}{k^{2}-k+1}-\displaystyle\sum_{n=1}^{2016}\frac{1}{k^{2}+k+1})$

$S=\frac{1}{2}(\displaystyle\sum_{n=1}^{2016}\frac{1}{k^{2}-k+1}-\displaystyle\sum_{n=2}^{2017}\frac{1}{k^{2}-k+1})$

$S=\frac{1}{2}(1-\frac{2017}{2017^{2}-2017+1})$

$S=\frac{1}{2}(\frac{2017(2016)}{2017(2016)+1})$

$S=\frac{2017(1008)}{2017(2016)+1}$

$\Rightarrow p+q=1008+2016=3024$