Question

坐标平面中的点列 $(a_1, b_1)$ ， $(a_2, b_2)$ ， $(a_3, b_3)$ ， $\ldots$ 满足 $\begin{aligned} &(a_{n+1}, b_{n+1}) \\ &= \left(\sqrt{3}\,a_n - b_n,\; \sqrt{3}\,b_n + a_n\right) \\ &\quad (n = 1, 2, 3, \ldots) \end{aligned}$ 已知 $(a_{100}, b_{100}) = (2, 4)$ 。求 $a_1 + b_1$ 。

A sequence $(a_1, b_1),$ $(a_2, b_2),$ $(a_3, b_3),$ $\ldots$ of points in the coordinate plane satisfies $\begin{aligned} &(a_{n+1}, b_{n+1}) \\ &= \left(\sqrt{3}\,a_n - b_n,\; \sqrt{3}\,b_n + a_n\right) \\ &\quad (n = 1, 2, 3, \ldots) \end{aligned}$ Suppose that $(a_{100}, b_{100}) = (2, 4).$ What is $a_1 + b_1?$

$-\dfrac{1}{2^{97}}$

$-\dfrac{1}{2^{99}}$

$0$

$\dfrac{1}{2^{98}}$

$\dfrac{1}{2^{96}}$

答案：D

知识点：复数棣莫弗定理

难度评级：2440

小提示：

写 $z_n = a_n + b_n i$ ；递推变为 $z_{n+1} = z_n(\sqrt{3} + i)$ 。

Write $z_n = a_n + b_n i;$ the recurrence becomes $z_{n+1} = z_n(\sqrt{3} + i)$

大提示：

因为 $\sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ)$ ，对 $(\sqrt{3} + i)^{99}$ 使用棣莫弗定理。

Since $\sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ),$ apply De Moivre’s theorem to $(\sqrt{3} + i)^{99}$

解答：

令 $z_n = a_n + b_n i$ ，则 $\begin{aligned} z_{n+1} &= (\sqrt{3}\,a_n - b_n) \\ &\quad {}+ (\sqrt{3}\,b_n + a_n)i \\ &= (a_n + b_n i)(\sqrt{3} + i)\text{，} \end{aligned}$ 所以 $z_{n+1} = z_n(\sqrt{3} + i)$ ，且 $z_{100} = z_1(\sqrt{3} + i)^{99}$ 。

因为 $\sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ)$ ，棣莫弗定理给出 $(\sqrt{3} + i)^{99}$ $= 2^{99}(\cos 2970^\circ + i\sin 2970^\circ)$ 。而 $2970^\circ$ 与 $90^\circ$ 同终边，所以它等于 $2^{99} i$ 。

于是 $2 + 4i = z_1 \cdot 2^{99} i$ ，所以 $z_1 = \dfrac{2 + 4i}{2^{99} i} = \dfrac{4 - 2i}{2^{99}}\text{。}$

因此 $a_1 = \tfrac{4}{2^{99}}$ ， $b_1 = -\tfrac{2}{2^{99}}$ ，所以 $a_1 + b_1 = \dfrac{2}{2^{99}} = \dfrac{1}{2^{98}}\text{。}$

所以正确答案是 D。

Let $z_n = a_n + b_n i.$ Then $\begin{aligned} z_{n+1} &= (\sqrt{3}\,a_n - b_n) \\ &\quad {}+ (\sqrt{3}\,b_n + a_n)i \\ &= (a_n + b_n i)(\sqrt{3} + i), \end{aligned}$ so $z_{n+1} = z_n(\sqrt{3} + i)$ and $z_{100} = z_1(\sqrt{3} + i)^{99}.$

Since $\sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ),$ De Moivre’s theorem gives $(\sqrt{3} + i)^{99}$ $= 2^{99}(\cos 2970^\circ + i\sin 2970^\circ).$ As $2970^\circ$ is coterminal with $90^\circ,$ this equals $2^{99} i.$

Thus $2 + 4i = z_1 \cdot 2^{99} i,$ so $z_1 = \dfrac{2 + 4i}{2^{99} i} = \dfrac{4 - 2i}{2^{99}}.$

Then $a_1 = \tfrac{4}{2^{99}}$ and $b_1 = -\tfrac{2}{2^{99}},$ so $a_1 + b_1 = \dfrac{2}{2^{99}} = \dfrac{1}{2^{98}}.$

Thus, D is the correct answer.

2008 AMC 12A 第 25 题

答案：D

难度评级：2440

小提示：

大提示：

解答：

其他年份的第 25 题