AMC 10 · 2006 · #1
Grade 7 arithmeticPick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
There are 2006 terms, so brute force is out. Tool #5 (Look for a Pattern) spots that the terms just alternate -1, +1, -1, +1, …. Tool #7 (Identify Subproblems) then chops the long sum into easy pairs, each of which collapses to 0. Tool #3 (Eliminate Possibilities) reads the choices: since (-1)ⁿ is only ever ± 1, the sum can never be as big as ± 2006, which kills (A) and (E) before any real work.
See what each term equals
Multiplying by -1 flips the sign every time, so (-1)ⁿ is -1 for odd n and +1 for even n.
Powers of -1 only ever land on -1 or +1, switching each time the exponent goes up by one.
6.EE.A.1Look For A PatternPair the terms two at a time
Pair each odd term with the even term right after it: every pair is (-1) + (+1) = 0.
A negative and its matching positive cancel, so each pair quietly disappears.
A negative and its matching positive cancel, so each pair quietly disappears.
▸ Why?
A quantity added to its own opposite leaves nothing behind.
▸ Why?
The sign flips with every step of the exponent, so the terms alternate and pair off exactly.
Count the pairs and finish
2006 is even, so the terms form 1003 whole pairs with none left over: the total is 0, choice (C).
An even count of terms pairs up perfectly, and a pile of zeros is still zero.
6.NS.B.2Eliminate PossibilitiesPowers of -1 just flip between -1 and +1, so pair them up — each pair cancels to zero, and an even number of terms leaves nothing behind.
- See what each term equals
- Pair the terms two at a time
- Count the pairs and finish