1964 AMC 12 Problem 34

Attempt Problem 34 of the 1964 AMC 12 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 1964 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

34.

If nn is a multiple of 4,4, the sum
s=1+2i+3i2+⋯+(n+1)in, \begin{aligned} s&=1+2i+3i^2+\cdots\\ &\quad+(n+1)i^n, \end{aligned}
where i=−1,i=\sqrt{-1}, equals:
1+i1+i
12(n+2)\dfrac12(n+2)
12(n+2−ni)\dfrac12(n+2-ni)
12[(n+1)(1−i)+2]\dfrac12\left[(n+1)(1-i)+2\right]
18(n2+8−4ni)\dfrac18(n^2+8-4ni)
Answer: C
Concepts:complex numberroots of unitysummation
Small hint:
Group the terms in blocks of four powers of ii
Big hint:
Each complete block beginning with coefficient 4j+14j+1 sums to −2−2i-2-2i
Solution:
Write n=4m.n=4m. For j=0,…,m−1,j=0,\ldots,m-1, the four terms with exponents 4j4j through 4j+34j+3 sum to (4j+1)+(4j+2)i−(4j+3)−(4j+4)i=−2−2i. \begin{aligned} &(4j+1)+(4j+2)i\\ &\quad-(4j+3)-(4j+4)i\\ &=-2-2i. \end{aligned} The final term is 4m+1.4m+1. Therefore s=m(−2−2i)+(4m+1)=n+2−ni2. \begin{aligned} s&=m(-2-2i)+(4m+1)\\ &=\frac{n+2-ni}{2}. \end{aligned}
Thus, the correct answer is C.
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