Taiwanese Mathematical Olympiad 2002

Given those real numbers, consider the rational polynomial

$\sum \frac{ a_i } { x + i } - \frac{ 4 } { 2x + 1 }$

If we clear denominators, we get the polynomial:

$\sum \left( (2x+1) a_i \prod_{j \neq i } ( x + j ) \right) - 4 \prod (x + j )$

which is a degree at most 2002 polynomial. Since it has at most 2002 roots, and we already found 2002 roots of the form $x = 1, 2, \ldots, 2002$ , thus we have the identity.

$\sum \left( (2x+1) a_i \prod_{j \neq i } ( x + j ) \right) - 4 \prod (x + j ) = C (x-1)(x-2) \ldots ( x - 2002 )$

It remains to determine $C$ . Substituting in $x= - \frac{1}{2}$ , we get that $-4 \prod ( - \frac{1}{2} + j) = C \prod ( - \frac{1}{2} - j )$ , or that $C = \frac{ - 4 } { 4005 }$ .

Finally, substituting in $x= \frac{1}{2}$ , we get

$\sum \left( 2 a_i \prod_{j \neq i } ( \frac{1}{2} + j ) \right) - 4 \prod (\frac{1}{2} + j ) = \frac{-4}{4005} (\frac{1}{2}-1)(\frac{1}{2}-2) \ldots (\frac{1}{2} - 2002 )$

Dividing throughout by $\prod ( \frac{1}{2} + j )$ , we obtain

$\sum \frac{ a_i}{ \frac{1}{2} + i } = 2 - \frac{2}{ 2005^2 }$

Dividing by 2, we thus get

$\frac{a_1}{ 3} + \frac{a_2} { 5 } + \ldots + \frac{a_{2002} } { 4005 } = 1 - \frac{1} { 2005^2 } = \frac{ 4020024 } { 4020025 }$