Messy quadratic

Let $\displaystyle A= a_1 +\sum_{i=1}^{2012} \frac{a_{i+1}^3}{a_i^2+a_ia_{i+1}+a_{i+1}^2} = a_1 +\sum_{i=1}^{2012} \frac{a_{i+1}^3-a_i^3+a_i^3}{a_i^2+a_ia_{i+1}+a_{i+1}^2}$

$\displaystyle = a_1 +\sum_{i=1}^{2012} (a_{i+1} -a_i) +\sum_{i=1}^{2012} \frac{a_i^3}{a_i^2+a_ia_{i+1}+a_{i+1}^2}$

The first sum is solved using telescoping series property to get $\displaystyle -a_1+a_{2013}$ .

We have $\displaystyle f(x)=\frac{-x+x\sqrt{4x-3}}{2}$ and $\displaystyle a_{i+1}=f(a_i)$ .

$\displaystyle x^2 +xf(x) +f(x)^2 = x^2 +f(x) (x+f(x))= x^2 +f(x) (x+ \frac{-x+x\sqrt{4x-3}}{2})$

$\displaystyle =x^2+ \frac{-x+x\sqrt{4x-3}}{2} \frac{x+x\sqrt{4x-3}}{2}$

$\displaystyle = x^2 + \frac{-4x^2 +4x^3}{4} = x^3$ $\displaystyle => a_i^2+a_ia_{i+1}+a_{i+1}^2 = a_i^3$ (let $\displaystyle x=a_i)$

$\displaystyle => A= a_1-a_1+a_{2013} +\sum_{i=1}^{2012} 1 = 2013+2012=\boxed{4025}$

Verify that $a_i^3 = a_i^2 + a_ia_{i + 1} + a_{i + 1}^2$ , so

$\displaystyle\sum_{i = 1}^{2012}\frac{a_{i + 1}^3}{a_i^2 + a_ia_{i + 1} + a_{i + 1}^2} - \displaystyle\sum_{i = 1}^{2012}\frac{a_{i }^3}{a_i^2 + a_ia_{i + 1} + a_{i + 1}^2} = \displaystyle\sum_{i = 1}^{2012}(a_{i + 1} - a_i) = a_{2013} - a_{1}.$

Since $\displaystyle\sum_{i = 1}^{2012}\frac{a_{i }^3}{a_i^2 + a_ia_{i + 1} + a_{i + 1}^2} = 2012$ , the desired answer is thus $2013 + 2012 = \boxed{4025}$ .