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: | + | 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>. | + | 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 x to be <math>7</math>, so | + | 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> | + | <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 10:55, 20 December 2018
Problem
The coefficient of 
in the expansion of 
is:
Solution
By the Binomial Theorem, each term of the expansion is 
.
We want the exponent of 
to be 
, so
![\[2(8-n)-n=7\]](http://latex.artofproblemsolving.com/d/a/4/da406a3c74e21d3222f3da34ca29c9499aafecb5.png)
![\[16-3n=7\]](http://latex.artofproblemsolving.com/b/4/2/b42cac801908ead0872f3b1a9bbdba75b038447a.png)
![\[n=3\]](http://latex.artofproblemsolving.com/7/4/a/74ac5915a3e58b5ed0da41429618c1f8ab34e598.png)
If 
, then the corresponding term is
![\[\binom{8}{3}\left(\frac{x^2}{2}\right)^{5}\left(\frac{-2}{x}\right)^3\]](http://latex.artofproblemsolving.com/7/9/2/79276dde4a859e87568d37980d5743845f2abf8f.png)
![\[56 \cdot \frac{x^{10}}{32} \cdot \frac{-8}{x^3}\]](http://latex.artofproblemsolving.com/2/2/6/2260d96bd30607a4f60bb7ac1bc0ee81668d6823.png)
![\[-14x^7\]](http://latex.artofproblemsolving.com/2/8/9/289a7c85f9e76ff317826b8a1cc3cada153d8e60.png)
The answer is 
.
See Also
| 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 | ||
