Difference between revisions of "1962 AHSME Problems/Problem 12"

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By the Binomial Theorem, the sum of the last three coefficients is  
 
By the Binomial Theorem, the sum of the last three coefficients is  
 
<math>\binom{6}{2}-\binom{6}{1}+\binom{6}{0}=15-6+1=\boxed{10 \textbf{ (C)}}</math>.
 
<math>\binom{6}{2}-\binom{6}{1}+\binom{6}{0}=15-6+1=\boxed{10 \textbf{ (C)}}</math>.
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==See Also==
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{{AHSME 40p box|year=1962|before=Problem 11|num-a=13}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:15, 3 October 2014

Problem

When $\left ( 1 - \frac{1}{a} \right ) ^6$ is expanded the sum of the last three coefficients is:

$\textbf{(A)}\ 22\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ -10\qquad\textbf{(E)}\ -11$

Solution

This is equivalent to $\frac{(a-1)^6}{a^6}.$ Its expansion has 7 terms, whose coefficients are the same as those of $(a-1)^6$. By the Binomial Theorem, the sum of the last three coefficients is $\binom{6}{2}-\binom{6}{1}+\binom{6}{0}=15-6+1=\boxed{10 \textbf{ (C)}}$.

See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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