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>. | + | 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>. |
− | + | ---- | |
− | * [[University of South Carolina High School Math Contest/1993 Exam]] | + | |
+ | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 7|Previous Problem]] | ||
+ | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 9|Next Problem]] | ||
+ | * [[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 in the expansion of
![$4 (1 + x + x^2 + x^3 + x^4 + x^5 )^6?$](http://latex.artofproblemsolving.com/9/5/0/950100c82ab65682e2e0bfd82ac80f7b3ef638f5.png)
![$\mathrm{(A) \ } 40 \qquad \mathrm{(B) \ }48 \qquad \mathrm{(C) \ }56 \qquad \mathrm{(D) \ }62 \qquad \mathrm{(E) \ } 64$](http://latex.artofproblemsolving.com/d/3/7/d37b89c36eabc473792f9ed081d97757a8f69f3d.png)
Solution
The expression simplifies to . Expanding both the numerator and denominator, we see that the coefficient of the
term is
.