Difference between revisions of "1960 AHSME Problems/Problem 20"

Revision as of 19:15, 10 May 2018

The coefficient of $x^7$ in the expansion of $(\frac{x^2}{2}-\frac{2}{x})^8$ is:

$\textbf{(A)}\ 56\qquad \textbf{(B)}\ -56\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ -14\qquad \textbf{(E)}\ 0$

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 $\boxed{\textbf{(D)}}$ .

1960 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AHSME Problems and Solutions

Revision as of 11:07, 10 May 2018 (view source) Rockmanex3 (talk \| contribs) (Solution to Problem 20)		Revision as of 19:15, 10 May 2018 (view source) Rockmanex3 (talk \| contribs) m (→‎See Also) Newer edit →
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	==See Also==		==See Also==
−	{{AHSME box\|year=1960\|num-b=19\|num-a=21}}	+	{{AHSME 40p box\|year=1960\|num-b=19\|num-a=21}}