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Difference between revisions of "1961 AHSME Problems/Problem 7"
(Solution to Problem 7)
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When simplified, the third term in the expansion of <math>(\frac{a}{\sqrt{x}}- \frac{\sqrt{x}}{a^2})^6</math> is:
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== Problem ==
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When simplified, the third term in the expansion of <math>(\frac{a}{\sqrt{x}}-\frac{\sqrt{x}}{a^2})^6</math> is:
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<math>\textbf{(A)}\ \frac{15}{x}\qquad
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\textbf{(B)}\ -\frac{15}{x}\qquad
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\textbf{(C)}\ -\frac{6x^2}{a^9} \qquad
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\textbf{(D)}\ \frac{20}{a^3}\qquad
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\textbf{(E)}\ -\frac{20}{a^3}</math>
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==Solution==
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By the [[Binomial Theorem]], the third term in the expansion is
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<cmath>\binom{6}{2}(\frac{a}{\sqrt{x}})^{4}(\frac{-\sqrt{x}}{a^2})^2</cmath>
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<cmath>15 \cdot \frac{a^4}{x^2} \cdot \frac{x}{a^4}</cmath>
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<cmath>\frac{15}{x}</cmath>
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The answer is <math>\boxed{\textbf{(A)}}</math>.
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==See Also==
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{{AHSME 40p box|year=1961|num-b=6|num-a=8}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:08, 17 May 2018

Problem

When simplified, the third term in the expansion of $(\frac{a}{\sqrt{x}}-\frac{\sqrt{x}}{a^2})^6$ is:

$\textbf{(A)}\ \frac{15}{x}\qquad \textbf{(B)}\ -\frac{15}{x}\qquad \textbf{(C)}\ -\frac{6x^2}{a^9} \qquad \textbf{(D)}\ \frac{20}{a^3}\qquad \textbf{(E)}\ -\frac{20}{a^3}$

Solution

By the Binomial Theorem, the third term in the expansion is \[\binom{6}{2}(\frac{a}{\sqrt{x}})^{4}(\frac{-\sqrt{x}}{a^2})^2\] \[15 \cdot \frac{a^4}{x^2} \cdot \frac{x}{a^4}\] \[\frac{15}{x}\] The answer is $\boxed{\textbf{(A)}}$.

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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