Question

表达式

$(x + y + z)^{2006} + (x - y - z)^{2006}$

展开并合并同类项后，化简后的表达式中有多少项？

The expression

$(x + y + z)^{2006} + (x - y - z)^{2006}$

is simplified by expanding it and combining like terms. How many terms are in the simplified expression?

$6018$

$671{,}676$

$1{,}007{,}514$

$1{,}008{,}016$

$2{,}015{,}028$

答案：D

知识点：二项式定理奇偶性基本计数

难度评级：2340

解答：

单项式 $x^a y^b z^c$ 只有在 $a$ 为偶数时保留，因为 $a$ 为奇数的项会在两个展开式之间抵消。

对每个满足 $0 \le a \le 2006$ 的偶数 $a$ ，指数 $b$ 有 $2007 - a$ 个取值，而 $c = 2006 - a - b$ 随后确定。对所有偶数 $a$ 求和：这是前 $1004$ 个正奇数之和，等于 $1004^2 = 1{,}008{,}016$ 。 $\begin{gathered} (2007 - 0) + (2007 - 2) \\ {}+ \cdots + (2007 - 2006) \\ = 2007 + 2005 + \cdots + 1, \end{gathered}$

因此，正确答案是 D。

A term $x^a y^b z^c$ survives only when $a$ is even, since terms with odd $a$ cancel between the two expansions.

For each even $a$ with $0 \le a \le 2006,$ the exponent $b$ ranges over $2007 - a$ values and $c = 2006 - a - b$ is then determined. Summing over even $a:$ $\begin{gathered} (2007 - 0) + (2007 - 2) \\ {}+ \cdots + (2007 - 2006) \\ = 2007 + 2005 + \cdots + 1, \end{gathered}$ the sum of the first $1004$ odd positive integers, which is $1004^2 = 1{,}008{,}016.$

Thus, the correct answer is D.

2006 AMC 12A 第 24 题

答案：D

难度评级：2340

解答：

其他年份的第 24 题