2002 AMC 10P Problem 23

Let a=\frac{1^2}{1} + \frac{2^2}{3} + \frac{3^2}{5} + \; \ldots \; + \frac{1001^2}{2001}

and

b=\frac{1^2}{3} + \frac{2^2}{5} + \frac{3^2}{7} + \; \ldots \; + \frac{1001^2}{2003}.

Find the integer closest to a-b.

\text{(A) }500 \qquad \text{(B) }501 \qquad \text{(C) }999 \qquad \text{(D) }1000 \qquad \text{(E) }1001

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.