AMC 10 · 2006 · #1
Grade 7 arithmeticWhat is (−1)1+(−1)2+...+(−1)2006 ?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Add up the powers $(-1)^1 + (-1)^2 + (-1)^3 + \cdots + (-1)^{2006}$, where the exponent runs through every whole number from $1$ to $2006$.
Givens: The sum $(-1)^1 + (-1)^2 + \cdots + (-1)^{2006}$; The base is $-1$ every time; only the exponent changes, running $1,2,3,\dots,2006$; Answer choices: (A) $-2006$, (B) $-1$, (C) $0$, (D) $1$, (E) $2006$
Unknowns: The single number the whole sum equals
Understand
Restated: Add up the powers $(-1)^1 + (-1)^2 + (-1)^3 + \cdots + (-1)^{2006}$, where the exponent runs through every whole number from $1$ to $2006$.
Givens: The sum $(-1)^1 + (-1)^2 + \cdots + (-1)^{2006}$; The base is $-1$ every time; only the exponent changes, running $1,2,3,\dots,2006$; Answer choices: (A) $-2006$, (B) $-1$, (C) $0$, (D) $1$, (E) $2006$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
There are $2006$ terms, so brute force is out. Tool #5 (Look for a Pattern) spots that the terms just alternate $-1, +1, -1, +1, \dots$. 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 $\pm 1$, the sum can never be as big as $\pm 2006$, which kills (A) and (E) before any real work.
Execute — Answer: C
6.EE.A.1 Step 1 See what each term equals
- The base $-1$ multiplied by itself flips sign every step.
- An odd number of factors leaves it negative and an even number makes it positive, so $(-1)^n = -1$ when $n$ is odd and $(-1)^n = +1$ when $n$ is even.
- The sum is therefore just $-1 + 1 - 1 + 1 - \cdots$, alternating all the way to the $2006$th term.
💡 Powers of $-1$ only ever land on $-1$ or $+1$, switching each time the exponent goes up by one.
7.NS.A.1 Step 2 Pair the terms two at a time
- Group the sum into consecutive pairs: an odd term with the even term right after it.
- Each pair is $(-1) + (+1) = 0$.
- For example $(-1)^1 + (-1)^2 = -1 + 1 = 0$, and the same thing happens for every following pair.
💡 A negative and its matching positive cancel, so each pair quietly disappears.
6.NS.B.2 Step 3 Count the pairs and finish
- There are $2006$ terms, and $2006$ is even, so they split into exactly $2006 \div 2 = 1003$ complete pairs with nothing left over.
- Every pair is $0$, so the total is $1003 \times 0 = 0$.
- That is choice (C).
- Because the terms only ever add $\pm 1$, the giant values (A) $-2006$ and (E) $2006$ are impossible, and the perfect even pairing leaves no stray $-1$ or $+1$, ruling out (B) and (D).
💡 An even count of terms pairs up perfectly, and a pile of zeros is still zero.
6.EE.A.1 The base $-1$ multiplied by itself flips sign every step. An odd number of facto 7.NS.A.1 Group the sum into consecutive pairs: an odd term with the even term right after 6.NS.B.2 There are $2006$ terms, and $2006$ is even, so they split into exactly $2006 \di Review
Reasonableness: Each term is only $-1$ or $+1$, so the running total can never stray far from $0$; that alone rules out (A) $-2006$ and (E) $2006$ immediately. Because $2006$ is even, the number of $-1$'s equals the number of $+1$'s (both $1003$), so they cancel exactly and no leftover $\pm 1$ survives — eliminating (B) and (D) and confirming $0$. A quick sanity check on a short version, $(-1)^1+(-1)^2=0$ and $(-1)^1+\cdots+(-1)^4=0$, shows every even-length sum lands on $0$.
Alternative: Count the two kinds of term separately. From $1$ to $2006$ there are $1003$ odd exponents (each giving $-1$) and $1003$ even exponents (each giving $+1$). The sum is $1003 \times (-1) + 1003 \times (+1) = -1003 + 1003 = 0$, the same answer (C).
CCSS standards used (min grade 7)
6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating each power $(-1)^n$ as $-1$ for odd $n$ and $+1$ for even $n$.)7.NS.A.1Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers (Adding each pair $(-1) + (+1) = 0$ and recognizing that the negatives and positives cancel.)6.NS.B.2Fluently divide multi-digit numbers using the standard algorithm (Splitting the $2006$ terms into $2006 \div 2 = 1003$ complete pairs.)
⭐ Powers 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.
⭐ Powers 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.
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