$\frac{1}{2}+\frac{2^{1}}{2^{2}}+\frac{2^{2}}{2^{3}}+\cdots+\frac{2^{2002}}{2^{2003}}+\frac{2^{2003}}{2^{2004}}$ is equal to
(A) 1002
(B) 501
(C) $\frac{1}{2^{2004}}$
(D) 2004
(E) $\frac{2004}{2^{2004}}$
Here we use the rule for manipulating exponents $\frac{2^{a}}{2^{b}}=2^{a-b}$.

Therefore, each of the 2003 fractions after the first fraction is equal to $2^{-1}=\frac{1}{2}$.

This gives us 2004 copies of $\frac{1}{2}$ being added up, for a total of 1002 .

Answer: (A)