1977 AMC 12 第 16 题

Question

若

i^2=-1

，则和

\begin{aligned} &\cos45^\circ+i\cos135^\circ+\cdots\\ &\quad+i^n\cos(45+90n)^\circ+\cdots\\ &\quad+i^{40}\cos3645^\circ \end{aligned}

等于

If

i^2=-1,

then the sum

\begin{aligned} &\cos45^\circ+i\cos135^\circ+\cdots\\ &\quad+i^n\cos(45+90n)^\circ+\cdots\\ &\quad+i^{40}\cos3645^\circ \end{aligned}

equals

\frac{\sqrt2}{2}

-10i\sqrt2

\frac{21\sqrt2}{2}

\frac{\sqrt2}{2}(21-20i)

\frac{\sqrt2}{2}(21+20i)

比较下标为

n+2

的项与下标为

n

的项

Compare the term indexed by

n+2

with the term indexed by

n

分别数出从

0

到

40

的偶数下标和奇数下标

Count the even and odd indices from

0

through

40

令

u_n=i^n\cos(45+90n)^\circ

。则

\begin{aligned} u_{n+2} &=i^{n+2}\cos(225+90n)^\circ\\ &=(-i^n)(-\cos(45+90n)^\circ)\\ &=u_n \end{aligned}\text{。}

每个偶数下标的项都等于

\frac{\sqrt2}{2}

，每个奇数下标的项都等于

-\frac{i\sqrt2}{2}

。共有

21

个偶数下标和

20

个奇数下标，所以总和为

\frac{\sqrt2}{2}(21-20i)

。

因此，正确答案是 D。

Let

u_n=i^n\cos(45+90n)^\circ.

Then

\begin{aligned} u_{n+2} &=i^{n+2}\cos(225+90n)^\circ\\ &=(-i^n)(-\cos(45+90n)^\circ)\\ &=u_n. \end{aligned}

Every even-indexed term equals

\frac{\sqrt2}{2},

and every odd-indexed term equals

-\frac{i\sqrt2}{2}.

There are

21

even indices and

20

odd indices, so the sum is

\frac{\sqrt2}{2}(21-20i).

Therefore, the correct answer is D.

其他年份的第 16 题