# What is the sum of the series S=1^2-2^2+3^2-4^2+...-2002^2+2003^2?

$2007006$
${\sum}_{k = 1}^{2003} {\left(- 1\right)}^{k + 1} {k}^{2} = {\sum}_{k = 1}^{1002} {\left(2 k - 1\right)}^{2} - {\sum}_{k = 1}^{1001} {\left(2 k\right)}^{2} =$
${\left(2 \cdot 1002 - 1\right)}^{2} + {\sum}_{k = 1}^{1001} \left({\left(2 k\right)}^{2} - {\left(2 k\right)}^{2} - 4 k + 1\right) =$
${2003}^{2} - {\sum}_{k = 1}^{1001} \left(4 k - 1\right) = {2003}^{2} - 4 \cdot \frac{1001 \left(1001 + 1\right)}{2} + 1001 = 2007006$