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!)^250!} \qquad \mathrm{(D) \ }100!\left(-\frac 1{50!50!} + \frac 1{25!75!} \qquad \mathrm{(E) \ } 0 </math></center>
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<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 ==

Revision as of 22:38, 21 July 2006

Problem

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

$\mathrm{(A) \ } -2^{50} \qquad \mathrm{(B) \ } 20^{50}  - \frac{100!}{50!50!} \qquad \mathrm{(C) \ } \frac{100!}{(25!)^2 50!} \qquad \mathrm{(D) \ } 100! \left(-\frac 1{50!50!} + \frac 1{25!75!}\right) \qquad \mathrm{(E) \ } 0$

Solution

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

See also