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Algebra Difficulty 5.9 AIME, harder Prove it Ukraine

Assume that positive real numbers a1,a2,,a2006a_1, a_2, \dots, a_{2006} satisfy the equality
a1(a1+a2)a2+a2(a2+a3)a3++a2005(a2005+a2006)a2006+a2006(a2006+a1)a1=2006 \frac{a_1}{(a_1+a_2)a_2} + \frac{a_2}{(a_2+a_3)a_3} + \dots + \frac{a_{2005}}{(a_{2005}+a_{2006})a_{2006}} + \frac{a_{2006}}{(a_{2006}+a_1)a_1} = 2006
Find the value of the expression
a1a2006(a2006+a1)+a2a1(a1+a2)++a2005a2004(a2004+a2005)+a2006a2005(a2005+a2006) \frac{a_1}{a_{2006}(a_{2006} + a_1)} + \frac{a_2}{a_1(a_1 + a_2)} + \dots + \frac{a_{2005}}{a_{2004}(a_{2004} + a_{2005})} + \frac{a_{2006}}{a_{2005}(a_{2005} + a_{2006})}

Solution

Відповідь: 2006.

Помітимо, що
ai(ai+ai+1)ai+1=1ai+11ai+ai+1,ai+1ai(ai+ai+1)=1ai1ai+ai+1 \frac{a_i}{(a_i + a_{i+1})a_{i+1}} = \frac{1}{a_{i+1}} - \frac{1}{a_i + a_{i+1}}, \quad \frac{a_{i+1}}{a_i(a_i + a_{i+1})} = \frac{1}{a_i} - \frac{1}{a_i + a_{i+1}}
де i=1,2006i=1,2006, a2007=a1a_{2007} = a_1. Додаванням цих рівностей одержимо, що значення сум, про які йдеться в умові задачі, будуть однаковими.

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