1960 AHSME Problems/Problem 20

Problem

The coefficient of $x^7$ in the expansion of $(\frac{x^2}{2}-\frac{2}{x})^8$ 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 $\boxed{\textbf{(D)}}$ .

1960 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1960 AHSME Problems/Problem 20

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