2006 AMC 12B Problem 1 — Solution | LIVE by Po-Shen Loh

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2006 AMC 12B Problem 1

Below is the professionally curated solution for Problem 1 of the 2006 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:exponent pairing and grouping

Difficulty rating: 840

1.

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

$-2006$

$-1$

$0$

$1$

$2006$

Solution:

Since $(-1)^k = -1$ for odd $k$ and $(-1)^k = 1$ for even $k,$ the terms alternate $-1, 1, -1, 1, \ldots$

There are $2006$ terms, forming $1003$ pairs, each equal to $(-1) + 1 = 0.$ The total is $0.$

Thus, the correct answer is C.

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Problem 1 in Other Years

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