2006 AMC 10B Problem 1

1. Write out the first few terms to see the pattern:
$$(-1)^1 = -1,\quad (-1)^2 = 1,\quad (-1)^3 = -1,\quad (-1)^4 = 1.$$
An odd power of $-1$ is $-1$, and an even power is $1$. So the terms simply switch back and forth.
2. Group the terms two at a time, starting from the front: $(-1 + 1) + (-1 + 1) + \cdots$. Each group adds to $0$.
3. Check whether anything is left over. The exponents run from $1$ to $2006$, so there are $2006$ terms. That is an even number, so the terms fill exactly $2006 \div 2 = 1003$ groups, and the last group ends with $(-1)^{2006} = 1$.
4. Every group is $0$, so the whole sum is $0$.

The answer is $\boxed{\textbf{(C)}\ 0}$.

When the terms of a sum just switch between $-1$ and $1$, the only thing that matters is how many terms there are. An even count cancels completely; an odd count leaves one term (the first one, here $-1$). Count the terms before doing anything else.

An odd power of $-1$ equals $-1$ and an even power equals $1$, so the terms read
$$-1,\ 1,\ -1,\ 1,\ \ldots$$
Group them two at a time: $(-1 + 1) + (-1 + 1) + \cdots$. Each group is $0$. There are $2006$ terms, an even number, so they split evenly into $1003$ groups with nothing left over.

The total is therefore $0$.

The answer is $\boxed{\textbf{(C)}\ 0}$.