We can show by induction that $n+\frac{1}{2}\left([n-1]+\frac{1}{2}\left(\cdots+\frac{1}{2} \cdot 2\right) \cdots\right)=2(n-1)$. For $n=3$ we have $3+\frac{1}{2} \cdot 2=4$, giving the base case, and if the result holds for $n$, then $(n+1)+\frac{1}{2} 2(n-1)=2 n=2(n+1)-2$. Thus the claim holds, and now plug in $n=2002$. Alternate Solution: Expand the given expression as $2002+2001 / 2+2000 / 2^{2}+\cdots+2 / 2^{2000}$. Letting $S$ denote this sum, we have $S / 2=2002 / 2+2001 / 2^{2}+\cdots+2 / 2^{2001}$, so $S-S / 2=$ $2002-\left(1 / 2+1 / 4+\cdots+1 / 2^{2000}\right)-2 / 2^{2001}=2002-\left(1-1 / 2^{2000}\right)-1 / 2^{2000}=2001$, so $S=4002$.