Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 4"
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If <math>(1 + i)^{100}</math> is expanded and written in the form <math>a + bi</math> where <math>a</math> and <math>b</math> are real numbers, then <math>a =</math> | If <math>(1 + i)^{100}</math> is expanded and written in the form <math>a + bi</math> where <math>a</math> and <math>b</math> are real numbers, then <math>a =</math> | ||
− | <center><math> \mathrm{(A) \ }-2^{50} \qquad \mathrm{(B) \ }20^{50}-\frac{100!}{50!50!} \qquad \mathrm{(C) \ } \frac{100!}{(25!)^ | + | <center><math> \mathrm{(A) \ } -2^{50} \qquad \mathrm{(B) \ } 20^{50} - \frac{100!}{50!50!} \qquad \mathrm{(C) \ } \frac{100!}{(25!)^2 50!} \qquad \mathrm{(D) \ } 100! \left(-\frac 1{50!50!} + \frac 1{25!75!}\right) \qquad \mathrm{(E) \ } 0 </math></center> |
== Solution == | == Solution == | ||
Notice that <math>(1+i)^{2}=2i</math>. We then have <math>(2i)^{50}=-2^{50}</math>. | Notice that <math>(1+i)^{2}=2i</math>. We then have <math>(2i)^{50}=-2^{50}</math>. | ||
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+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 13:36, 24 November 2016
Problem
If is expanded and written in the form where and are real numbers, then
Solution
Notice that . We then have .