2006 AMC 12A Problem 24

Attempt Problem 24 of the 2006 AMC 12A 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 12A solutions, or check the answer key.

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24.

The expression
(x+y+z)2006+(x−y−z)2006 (x + y + z)^{2006} + (x - y - z)^{2006}
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
60186018
671,676671{,}676
1,007,5141{,}007{,}514
1,008,0161{,}008{,}016
2,015,0282{,}015{,}028
Answer: D
Concepts:binomial theoremparitybasic counting
Small hint:
A monomial xaybzcx^a y^b z^c survives only when aa is even; the odd-aa terms cancel
Big hint:
For each even aa from 00 to 2006,2006, count the possible values of bb
Solution:
A term xaybzcx^a y^b z^c survives only when aa is even, since terms with odd aa cancel between the two expansions.
For each even aa with 0≤a≤2006,0 \le a \le 2006, the exponent bb ranges over 2007−a2007 - a values and c=2006−a−bc = 2006 - a - b is then determined. Summing over even a:a: (2007−0)+(2007−2)+⋯+(2007−2006)=2007+2005+⋯+1, \begin{gathered} (2007 - 0) + (2007 - 2) \\ {}+ \cdots + (2007 - 2006) \\ = 2007 + 2005 + \cdots + 1, \end{gathered} the sum of the first 10041004 odd positive integers, which is 10042=1,008,016.1004^2 = 1{,}008{,}016.
Thus, the correct answer is D.
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