Sum of the series $$1^{2} - 2^{2} + 3^{2} - 4^{2} + ... + 2001^{2} - 2002^{2} + 2003^{2}$$ is:

The given series is $$1^2 - 2^2 + 3^2 - 4^2 +.....+2003^2$$
$$1^2 - 2^2$$ can be written as $$(1+2)(1-2)$$ = $$3*(-1)$$ = $$-3$$
$$3^2 -4^2$$ can be written as $$(3+4)*(3-4)$$ = $$7*(-1)$$ = $$-7$$
$$5^2-6^2$$ can be written as $$(5+6)*(5-6)$$ = $$11*(-1)$$ = $$-11$$
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 $$-3 - 7 ...-4003 + 2003^2$$
Let is evaluate the value of $$-3 - 7 ...-4003 $$
Number of terms,$$n$$ = $$\frac{4003-3}{4} + 1$$ = $$1001$$
Sum = $$\frac{n}{2} *$$(first term + last term)
       = $$\frac{1001}{2}*(-4006)$$
       = $$ -2005003$$
$$2003^2 = 4012009$$
Value of the given expression = $$4012009 - 2005003 = 2007006$$.

Therefore, option A is the right answer.