# Difference between revisions of "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 $\boxed{\textbf{(D)}}$.

## See Also

 1960 AHSC (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 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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