# 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 the x-term 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 $\boxed{\textbf{(D)}}$.