1960 AHSME Problems/Problem 20

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Problem

The coefficient of $x^7$ in the expansion of $\left(\frac{x^2}{2}-\frac{2}{x}\right)^8$ is:

$\textbf{(A)}\ 56\qquad \textbf{(B)}\ -56\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ -14\qquad \textbf{(E)}\ 0$

Solution

By the Binomial Theorem, each term of the expansion is $\binom{8}{n}(\frac{x^2}{2})^{8-n}(\frac{-2}{x})^n$.

We want the exponent of $x$ to be $7$, so \[2(8-n)-n=7\] \[16-3n=7\] \[n=3\]

If $n=3$, then the corresponding term is \[\binom{8}{3}(\frac{x^2}{2})^{5}(\frac{-2}{x})^3\] \[56 \cdot \frac{x^{10}}{32} \cdot \frac{-8}{x^3}\] \[-14x^7\]

The answer is $\boxed{\textbf{(D)}}$.

See Also

1960 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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