1960 AHSME Problems/Problem 20

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 $\binom{8}{n}\left(\frac{x^2}{2}\right)^{8-n}\left(\frac{-2}{x}\right)^n$.

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 \[\binom{8}{3}\left(\frac{x^2}{2}\right)^{5}\left(\frac{-2}{x}\right)^3\] \[56 \cdot \frac{x^{10}}{32} \cdot \frac{-8}{x^3}\] \[-14x^7\]

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

See Also

1960 AHSC (ProblemsAnswer KeyResources)
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