1961 AHSME Problems/Problem 7

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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