A year problem by Nihar

$\left(\dfrac{x(x+1)}{2}\right)^2 - \left(\dfrac{x(x-1)}{2}\right)^2=\dfrac{x^2}{2014}$

$\Rightarrow \left(\dfrac{x(x+1)}{2} + \dfrac{x(x-1)}{2}\right)\left(\dfrac{x(x+1)}{2} - \dfrac{x(x-1)}{2}\right)=\dfrac{x^2}{2014}$

$\Rightarrow \left(\dfrac{x^2+x+x^2-x}{2}\right)\left(\dfrac{x^2+x-x^2+x}{2}\right)=\dfrac{x^2}{2014}$

$\Rightarrow \left(\dfrac{2x^2}{2}\right)\left(\dfrac{2x}{2}\right)=\dfrac{x^2}{2014}$

$\Rightarrow x^3-\dfrac{x^2}{2014}=0$

$\Rightarrow x^2\left(x-\dfrac{1}{2014}\right)=0$

From this we get $x=\left(0,\dfrac{1}{2014}\right)$ , hence

$S_{2014}=\dfrac{1}{2014}=\dfrac{a}{b}$

$\huge\Rightarrow \boxed{\color{#3D99F6}{a+b=2015}}$

We also find that for this polynomial , $\huge S_n=\dfrac{1}{n}$

Just for a change one can see this as sum of cubes of first x terms minus sum of cubes of first x-1 terms. This directly reduces the expression to $x^3=\frac{x^2}{2014}$

$\large \left(\dfrac{x(x+1)}{2}\right)^2 - \left(\dfrac{x(x-1)}{2}\right)^2=\dfrac{x^2}{2014}\\\large \left(\dfrac{x^2}{4}\right)*\{(x+1)^2-(x-1)^2\}=\dfrac{x^2}{4}*\dfrac{4}{2014}\\\implies \large { {2x*2}=\dfrac{4}{2014} }~~~~~~\because (a^2-b^2)=(a+b)(a-b)\\x=\dfrac{1}{2014}=\dfrac{a}{b}.~~~~\therefore a+b= \large \color{#D61F06}{2015}$