# Difference between revisions of "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 $\boxed{\textbf{(C)}}$.

## See Also

 1964 AHSC (Problems • Answer Key • Resources) Preceded byProblem 33 Followed byProblem 35 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions

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