Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Prove it Ukraine

Let $f(x) = \frac{x^3}{3x^2 - 3x + 1}$ . Compute
$f\left(\frac{1}{2012}\right) + f\left(\frac{2}{2012}\right) + \dots + f\left(\frac{2012}{2012}\right).$

Solution

Зауважимо, що $f\left(\frac{i}{2012}\right) = \frac{i^3}{i^3 + (2012 - i)^3}$ , $1 \le i \le 2012$ . Тоді
$f\left(\frac{1006}{2012}\right) = \frac{1}{2}, \quad f\left(\frac{2012}{2012}\right) = 1, \quad f\left(\frac{i}{2012}\right) + f\left(\frac{2012 - i}{2012}\right) = 1, \quad 1 \le i \le 1005.$
Відповідь: $\frac{2013}{2}$ .

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