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

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2006 AMC 12B Problem 1

Attempt Problem 1 of the 2006 AMC 12B below, then check your answer against the professionally curated solution 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 problems are used with official legal permission of the Mathematical Association of America (MAA).

1.

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

$-2006$

$-1$

$0$

$1$

$2006$

Answer: C

Concepts:exponent pairing and grouping

Difficulty rating: 840

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.

Problem 2#2 →

Problem 1 in Other Years

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