<s>[INST] How can I determine the value of $a_1 + \displaystyle\sum_{i = 1}^{2012}\frac{a_{i + 1}^3}{a_i^2 + a_ia_{i + 1} + a_{i + 1}^2}$ For reals $x \ge 3$, let $f(x)$ denote the function $f(x) = \frac{-x + x\sqrt{4x - 3}}{2}$. Now suppose that $a_1, a_2, \ldots, a_{2013}$ is a sequence of real numbers such that $a_1 > 3, a_{2013} = 2013$, and for $n = 1, 2, \ldots, 2012$, $a_{n + 1} = f(a_n)$.
Given this, how can I determine the value of $$a_1 + \displaystyle\sum_{i = 1}^{2012}\frac{a_{i + 1}^3}{a_i^2 + a_ia_{i + 1} + a_{i + 1}^2}$$
 [/INST] 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 $\displaystyle -a_1+a_{2013}$.
Now we have $\displaystyle f(x)=\frac{-x+x\sqrt{4x-3}}{2}$ and $\displaystyle a_{i+1}=f(a_i)$
$$\displaystyle x^2 +xf(x) +f(x)^2 = x^2 +f(x) (x+f(x))= x^2 +f(x) \left (x+ \frac{-x+x\sqrt{4x-3}}{2}\right )$$
$$\displaystyle =x^2+ \frac{-x+x\sqrt{4x-3}}{2} \frac{x+x\sqrt{4x-3}}{2} $$
$$\displaystyle = x^2 + \frac{-4x^2 +4x^3}{4} = x^3$$ $$\displaystyle \Rightarrow a_i^2+a_ia_{i+1}+a_{i+1}^2 = a_i^3 \quad\text{(let } \displaystyle x=a_i\text{)}$$
$$\displaystyle \Rightarrow A= a_1-a_1+a_{2013} +\sum_{i=1}^{2012} 1 = 2013+2012=4025$$
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