2006 AMC 10B Problem 1 — Solution | LIVE by Po-Shen Loh

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2006 AMC 10B Problem 1

Attempt Problem 1 of the 2006 AMC 10B 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 10B 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: 720

Solution:

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.

Problem 2#2 →

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