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

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== Solution ==
 
== Solution ==
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>.
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The expression simplifies to <math>\left(\frac{x^{6}-1}{x-1}\right)^{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 ==
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* [[University of South Carolina High School Math Contest/1993 Exam]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 7|Previous Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 9|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 12:27, 23 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 $\left(\frac{x^{6}-1}{x-1}\right)^{6}$. Expanding both the numerator and denominator, we see that the coefficient of the $x^{3}$ term is ${6\choose 5}+{6\choose 3}+{6\choose 6}+{6\choose 3}=56$.