Indian Institute of Foreign Trade 2008 Solved Paper

The given series is $1^2-2^2+3^2-4^2+.....+{2003}^2$
$1^2-2^2$ can be written as $\left(1+2\right)\left(1-2\right)-3\ast \left(-1\right)=-3$
$3^2-4^2$ can be written as $\left(3+4\right)\ast \left(3-4\right)=7\ast \left(-1\right)=-7$
$5^2-6^2$ can be written as $\left(5+6\right)\ast \left(5-6\right)=11\ast \left(-1\right)=-11$
Therefore, all the terms till ${2002}^2$ can be expressed as an AP.
The last term of the AP will be $\left(2001+2002\right)\left(2001-2002\right)=-4003$
Therefore, the given expression is reduced to $-3-7...-4003+{2003}^2$
Let is evaluate the value of $-3-7...-4003$
Number of terms,$\mathit{n}=\frac{4003-3}{4}+1=1001$
Sum$=\frac{\mathit{n}}{2}\ast$(first term + last term)
$=\frac{1001}{2}\ast \left(-4006\right)$
$=-2005003$
${2003}^2=4012009$
Value of the given expression $=4012009-2005003=2007006$
Therefore, option A is the right answer.