AMC 10 · 2006 · #1

Grade 7 arithmetic
exponentsparitypattern-recognition identify-subproblems ↑ Prerequisites: exponentsparity
📏 Short solution 💡 1 insight
Problem
The powers of negative one are added up, with the exponent running from 1 to 2006. Find the total.

Pick an answer.

(A)
- 2006
(B)
- 1
(C)
0
(D)
1
(E)
2006
How to solve
Strategy Look for a Pattern

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.

1STEP 1

See what each term equals

The terms alternate between minus one and plus one.

(-1)¹=-1, (-1)²=+1, (-1)³=-1, (-1)⁴=+1, …
2STEP 2

Pair the terms two at a time

Pairing neighbours makes every pair zero.

(-1+1)₀ + (-1+1)₀ + … = 0 + 0 + …
3STEP 3

Count the pairs and finish

The count is even, so nothing is left over and the total is 0, choice (C).

2006 ÷ 2 = 1003 pairs, 1003 × 0 = 0 → (C)
Answer
0
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 ± 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+…+(-1)⁴=0, shows every even-length sum lands on 0.
💡Key takeaway

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.

  • See what each term equals
  • Pair the terms two at a time
  • Count the pairs and finish