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University of South Carolina High School Math Contest/1993 Exam/Problem 4

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Problem

If $(1 + i)^{100}$ is expanded and written in the form $a + bi$ where $a$ and $b$ are real numbers, then $a =$

$\mathrm{(A) \ } -2^{50} \qquad \mathrm{(B) \ } 20^{50} - \frac{100!}{50!50!} \qquad \mathrm{(C) \ } \frac{100!}{(25!)^2 50!}...$

Solution

Notice that $(1+i)^{2}=2i$ . We then have $(2i)^{50}=-2^{50}$ .

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/University_of_South_Carolina_High_School_Math_Contest/1993_Exam/Problem_4"

Category: Intermediate Algebra Problems

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