Complicated Expression

First, note that both solutions of $x^2+x+2014=0$ have a magnitude of $\sqrt{2014}$ and a irrational argument. This means that as $n\to \infty$ , the values of $\dfrac{x^i}{2014^{\frac{i-1}{2}}}+1$ where $1\le i\le n$ become the set of points with distance $\sqrt{2014}$ away from $1$ on the complex plane (because $\left|\dfrac{x^i}{2014^{\frac{i-1}{2}}}\right|=\sqrt{2014}$

Thus, $\left|\dfrac{x^i}{2014^{\frac{i-1}{2}}}+1\right|$ are the distances between the circle of radius $\sqrt{2014}$ and center $1$ and the origin.

Also, $\displaystyle\sqrt[\Large n]{\prod_{i=1}^{2n}\text{stuff}}=\sqrt[\Large \frac{n}{2}]{\prod_{i=1}^{n}\text{stuff}}=\sqrt[\Large n]{\prod_{i=1}^{n}\text{stuff}}^2$

Thus we just need to find the square of the geometric mean of all the distances between the origin and the circle.

Now given any point $P$ on the circle, draw a line through that point and $1$ . The line will cross the circle at another point $Q$ . If we let $1$ be point $X$ , then by Power of a Point, $(PX)(QX)=(\sqrt{2014}+1)(\sqrt{2014}-1)=2013$ .

However, we can apply this strategy for any point $P$ on the circle, and since the geometric mean of any two opposite distances on the circle (the $PX$ and the $QX$ ) is always $\sqrt{(PX)(QX)}=\sqrt{2013}$ , then the geometric mean of the entire thing is also $\sqrt{2013}$ .

However, we wanted the square of the geometric mean, so our answer is $\sqrt{2013}^2=\boxed{2013}$ .

I have a feeling in my gut that this answer is wrong and we need calculus.