Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 8"

 
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== Problem ==
 
== Problem ==
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What is the coefficient of <math>x^3</math> in the expansion of
  
<center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ } </math></center>
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<center><math>4 (1 + x + x^2 + x^3 + x^4 + x^5 )^6? </math></center>
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<center><math> \mathrm{(A) \ } 40 \qquad \mathrm{(B) \ }48 \qquad \mathrm{(C) \ }56 \qquad \mathrm{(D) \ }62 \qquad \mathrm{(E) \ } 64 </math></center>
  
 
== Solution ==
 
== Solution ==
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The expression simplifies to <math>(\frac{x^{6}-1}{x-1})^{6}</math>. Expanding both the numerator and denominator, we see that the coefficient of the <math>x^{3}</math> term is <math>{6\choose 5}+{6\choose 3}+{6\choose 6}+{6\choose 3}=56</math>.
  
 
== See also ==
 
== See also ==
 
* [[University of South Carolina High School Math Contest/1993 Exam]]
 
* [[University of South Carolina High School Math Contest/1993 Exam]]
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[[Category:Introductory Combinatorics Problems]]
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[[Category:Introductory Algebra Problems]]

Revision as of 20:40, 22 July 2006

Problem

What is the coefficient of $x^3$ in the expansion of

$4 (1 + x + x^2 + x^3 + x^4 + x^5 )^6?$
$\mathrm{(A) \ } 40 \qquad \mathrm{(B) \ }48 \qquad \mathrm{(C) \ }56 \qquad \mathrm{(D) \ }62 \qquad \mathrm{(E) \ } 64$

Solution

The expression simplifies to $(\frac{x^{6}-1}{x-1})^{6}$. Expanding both the numerator and denominator, we see that the coefficient of the $x^{3}$ term is ${6\choose 5}+{6\choose 3}+{6\choose 6}+{6\choose 3}=56$.

See also