Supersized Sums

Relevant wiki: Sum of n, n², or n³

$A=\dfrac{\displaystyle \sum^{2016}_{r=0}(6r^{2})-2016^{2}-2016}{(2016^{2}+1)^{2}-(2015\times 2017)^{2}}$

$A=\dfrac{6(\color{#D61F06}{1^2+2^2+3^2+...+2016^2})-2016^2-2016}{(2016^2+1)^2-(2015×2017)^2}$

Let $a=2016$ .

$A=\dfrac{6\left(\color{#D61F06}{\frac{a×(a+1)×(2a+1)}{6}}\right)-a^2-a}{(a^2+1)^2 -[(a-1)(a+1)]^2}$

$A=\dfrac{2a^3+3a^2+a-a^2-a}{a^4+1+2a^2-a^4-1+2a^4}$

$A=\dfrac{2a^2(a+1)}{4a^2}$

$A=\dfrac{a+1}{2}$

$\therefore A=\dfrac{2016+1}{2}=\boxed{1008.5}$

Same solution as @William Whitehouse 's presented as follows.

Let $n=2016$ , then we have:

$\begin{aligned} A & = \frac {\displaystyle 6 \sum_{r=0}^n r^2 - n^2 -n}{(n^2+1)^2-(n-1)^2(n+1)^2} \\ & = \frac {6\frac{n(n+1)(2n+1)}6 - n(n+1)}{(n^2+1)^2-(n^2-1)^2} \\ & = \frac {n(n+1)(2n+1-1)}{n^4+2n^2+1-(n^4-2n^2+1)} \\ & = \frac {2n^2(n+1)}{4n^2} \\ & = \frac {n+1}2 & \small \color{#3D99F6}{\text{Putting back }n = 2016} \\ & = \frac {2016+1}2 = \boxed{1008.5} \end{aligned}$

Let n=2016. We now have $\frac{n(n+1)(2n+1)-(n^{2}+n)}{(n^{2}+1)^{2}-((n-1)(n+1))^2}$ .

The top clearly resolves to $2n^{2}(n+1)$ and the bottom becomes $(n^{2}+1)^{2}-(n^2-1)^{2} = 4n^{2}$ this simplifies to $\frac{n+1}{2}$ which is $\frac{2017}{2} = \underline{1008.5}$