2003 AMC 10B Problem 1

1. Look at the top and bottom rows one position at a time: $2$ sits over $3$, $4$ over $6$, $6$ over $9$, and so on up to $14$ over $21$. Each top number is $2$ times something and the bottom number is $3$ times the same thing.
2. So the top row is $2$ times $(1-2+3-4+5-6+7)$ and the bottom row is $3$ times the same thing:
$$\frac{2\,(1-2+3-4+5-6+7)}{3\,(1-2+3-4+5-6+7)}.$$
3. The bracket is not zero: grouping it as $(1-2)+(3-4)+(5-6)+7 = -1-1-1+7 = 4$. Since it is a nonzero number on top and bottom, it cancels, leaving $\frac{2}{3}$.
4. Check by adding directly. Top: $(2-4)+(6-8)+(10-12)+14 = -2-2-2+14 = 8$. Bottom: $(3-6)+(9-12)+(15-18)+21 = -3-3-3+21 = 12$. And $\frac{8}{12} = \frac{2}{3}$.

The answer is $\boxed{\textbf{(C)}\ \frac{2}{3}}$.

The top row is just the bottom row shrunk by the same factor, term by term. When every piece of one sum is the same multiple of the matching piece of another sum, the whole sums have that same ratio. Before adding a long row of numbers, check whether it is a copy of another row times a constant.