Taiwanese Mathematical Olympiad 2002

Algebra Level 5

a12+a23++a20022003=43a13+a24++a20022004=45a12003+a22004++a20024004=44005\begin{aligned} \dfrac{a_{1}}{2} + \dfrac{a_{2}}{3} + \cdots + \dfrac{a_{2002}}{2003} &= \dfrac{4}{3}\\\\ \dfrac{a_{1}}{3} + \dfrac{a_{2}}{4} + \cdots + \dfrac{a_{2002}}{2004} &= \dfrac{4}{5}\\ & \vdots\\ \dfrac{a_{1}}{2003} + \dfrac{a_{2}}{2004} + \cdots + \dfrac{a_{2002}}{4004} &= \dfrac{4}{4005} \end{aligned}

Given that real numbers a1,a2,,a2002a_1, a_2, \ldots, a_{2002} satisfy the system of equations above, evaluate the following sum: a13+a25++a20024005.\dfrac{a_1}{3} + \dfrac{a_2}{5} + \cdots + \dfrac{a_{2002}}{4005}.

If your answer is of the form mn\dfrac{m}{n}, where mm and nn are coprime positive integers, submit m+nm+n.

×

Problem Loading...

Note Loading...

Set Loading...