The given series is 12−22+32−42+.....+20032 12−22 can be written as (1+2)(1−2)−3∗(−1)=−3 32−42 can be written as (3+4)∗(3−4)=7∗(−1)=−7 52−62 can be written as (5+6)∗(5−6)=11∗(−1)=−11 Therefore, all the terms till 20022 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...−4003+20032 Let is evaluate the value of −3−7...−4003 Number of terms,n=
4003−3
4
+1=1001 Sum=
n
2
∗(first term + last term) =
1001
2
∗(−4006) =−2005003 20032=4012009 Value of the given expression =4012009−2005003=2007006 Therefore, option A is the right answer.