2006 AMC 10B Problem 1

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 $\boxed{\textbf{(C)}\ 0}$.

An alternating sum of equal-magnitude terms depends only on whether the number of terms is even or odd: even means everything cancels, odd means one term survives. Check the parity of the term count first; that decides the answer before any arithmetic.

Geometric series with first term $-1$, ratio $-1$, and $2006$ terms: the sum is $\frac{-1\,(1 - (-1)^{2006})}{1 - (-1)} = \frac{-1 \cdot 0}{2} = 0$.

• Miscounting the number of terms and leaving an unpaired term, which gives -1 or 1.
• Reading the sum as $2006$ copies of $-1$ or of $1$ and choosing (A) or (E).