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

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==Problem==
 
==Problem==
  
The coefficient of <math>x^7</math> in the expansion of <math>(\frac{x^2}{2}-\frac{2}{x})^8</math> is:
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The coefficient of <math>x^7</math> in the expansion of <math>\left(\frac{x^2}{2}-\frac{2}{x}\right)^8</math> is:
  
 
<math>\textbf{(A)}\ 56\qquad
 
<math>\textbf{(A)}\ 56\qquad
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==Solution==
 
==Solution==
By the Binomial Theorem, each term of the expansion is <math>\binom{8}{n}(\frac{x^2}{2})^{8-n}(\frac{2}{x})^n</math>.
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By the [[Binomial Theorem]], each term of the expansion is <math>\binom{8}{n}\left(\frac{x^2}{2}\right)^{8-n}\left(\frac{-2}{x}\right)^n</math>.
  
We want the exponent of the x-term to be <math>7</math>, so  
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We want the exponent of <math>x</math> to be <math>7</math>, so  
 
<cmath>2(8-n)-n=7</cmath>
 
<cmath>2(8-n)-n=7</cmath>
 
<cmath>16-3n=7</cmath>
 
<cmath>16-3n=7</cmath>
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If <math>n=3</math>, then the corresponding term is
 
If <math>n=3</math>, then the corresponding term is
<cmath>\binom{8}{3}(\frac{x^2}{2})^{5}(\frac{-2}{x})^3</cmath>
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<cmath>\binom{8}{3}\left(\frac{x^2}{2}\right)^{5}\left(\frac{-2}{x}\right)^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>
 
<cmath>-14x^7</cmath>
 
<cmath>-14x^7</cmath>
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==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1960|num-b=19|num-a=21}}
 
{{AHSME 40p box|year=1960|num-b=19|num-a=21}}
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[[Category:Intermediate Algebra Problems]]

Latest revision as of 11:55, 20 December 2018

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
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