2002 AMC 12P Problems/Problem 14

Revision as of 18:27, 10 March 2024 by The 76923th (talk | contribs) (Solution)

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 \[4 + 8 + ... + 2020\] 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 \[-(2 + 6 + ... + 2022)\] 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 \[(1 + 5 + ... + 2021)i\]

See also

2002 AMC 12P (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 12 Problems and Solutions

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