If i^2=-1, then the sum &cos45°+icos135°+… & +i^ncos(45+90n)°+… & +i^40cos3645° equals

1977 AMC 12 Problem 16

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16.

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)

Compare the term indexed by

n+2

with the term indexed by

n

Count the even and odd indices from

0

through

40

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.