2002 AMC 12P Problems/Problem 14

Problem

Find $i + 2i^2 +3i^3 + ... + 2002i^{2002}.$

$\text{(A) }-999 + 1002i \qquad \text{(B) }-1002 + 999i \qquad \text{(C) }-1001 + 1000i \qquad \text{(D) }-1002 + 1001i \qquad \text{(E) }i$

Solution

Note that $i^4 = 1$ , so $i^n = i^{4m+n}$ for all integers $m$ and $n$ . In particular, $i = 1$ , $i^2 = -1$ , and $i^3 = -i$ . We group the positive and negative real terms together and group the positive and negative imaginary parts together. The positive real terms have exponents on $i$ that are multiples of 4. Therefore, the positive real part evaluates to $4 + 8 + ... + 2000$ The negative real terms have exponents on $i$ that are of the form $4k + 2$ for integers $k$ . Therefore, the negative real part evaluates to $-(2 + 6 + ... + 2002)$ The positive imaginary terms have exponents on $i$ that are of the form $4k + 1$ for integers $k$ . Therefore, the negative real part evaluates to $(1 + 5 + ... + 2001)i$ The negative imaginary terms have exponents on $i$ that are of the form $4k + 3$ for integers $k$ . Therefore, the negative real part evaluates to $-(3 + 7 + ... + 1999)i$

Putting everything together, we have

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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
All AMC 12 Problems and Solutions

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2002 AMC 12P Problems/Problem 14

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