Indian Institute of Foreign Trade 2008 Solved Paper

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