2006 AMC 10B 第 1 题

Question

求

(-1)^1 + (-1)^2 + \cdots + (-1)^{2006}

的值。

What is

(-1)^1 + (-1)^2 + \cdots + (-1)^{2006}?

-2006

-1

0

1

2006

将相邻两项分组。

Group the terms in consecutive pairs

每组中的奇次幂项和偶次幂项符号相反。

In each pair, compare the signs of the odd-power and even-power terms

2006

项。把相邻项配对，得到

(-1+1)+(-1+1)+\cdots

。由于

2006

是偶数，每一项都能配对，因此总和为

0

。

所以正确答案是 C。

There are

2006

terms. Pairing consecutive terms gives

(-1+1)+(-1+1)+\cdots.

Since

2006

is even, every term pairs off and the sum is

0.

Thus, the correct answer is C.

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