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

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m (Solution)
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<cmath>n=3</cmath>
 
<cmath>n=3</cmath>
  
If n=3, then the corresponding term is
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If <math>n=3</math>, then the corresponding term is
 
<cmath>\binom{8}{3}(\frac{x^2}{2})^{5}(\frac{-2}{x})^3</cmath>
 
<cmath>\binom{8}{3}(\frac{x^2}{2})^{5}(\frac{-2}{x})^3</cmath>
 
<cmath>56 \cdot \frac{x^{10}}{32} \cdot \frac{-8}{x^3}</cmath>
 
<cmath>56 \cdot \frac{x^{10}}{32} \cdot \frac{-8}{x^3}</cmath>

Revision as of 11:04, 11 May 2018

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

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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