1964 AHSME Problems/Problem 34

Problem

If $n$ is a multiple of $4$ , the sum $s=1+2i+3i^2+\cdots+(n+1)i^n$ , where $i=\sqrt{-1}$ , equals:

$\textbf{(A) }1+i\qquad\textbf{(B) }\frac{1}{2}(n+2)\qquad\textbf{(C) }\frac{1}{2}(n+2-ni)\qquad$

$\textbf{(D) }\frac{1}{2}[(n+1)(1-i)+2]\qquad \textbf{(E) }\frac{1}{8}(n^2+8-4ni)$

Solution

The real part is $1 + 3i^2 + 5i^4 + ...$ , which is $1 - 3 + 5 - 7 + ...$ . If $n$ is a multiple of $4$ , then we have an odd number of terms in total: we start with $1$ at $n=0$ , then add two more terms to get $1 - 3 + 5$ at $n=4$ , etc. With each successive addition, we're really adding a total of $2$ , since $-3 + 5 = 2$ , and $-7 + 9 = 2$ , etc.

At $n = 0$ , the sum is $1$ , and at $n=4$ , the sum is $3$ . Since the sum increases linearly, the real part of the sum is $1 + \frac{1}{2}n$ .

The imaginary part is $\frac{1}{i}(2i + 4i^3 + 6i^5 + ...)$ , which is $2 - 4 + 6 - 8 + ...$ . This time, we have an even number of terms. We group pairs of terms to get $(2-4) + (6-8) + ...$ , and notice that each pair gives $-2$ . Again, with $n=0$ the imaginary part is $0$ , while with $n=4$ the imaginary part is $-2$ . Since again the sum increases linearly, this means the imaginary part is $-\frac{n}{2}$ .

Combining the real and imaginary parts gives $\frac{n}{2} + 1 - \frac{n}{2}i$ , which is equivalent to option $\textbf{(C)}$ .

1964 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 33	Followed by Problem 35
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1964 AHSME Problems/Problem 34

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