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
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

Page

Toolbox

Search

1960 AHSME Problems/Problem 20

Problem

Solution

See Also