Difference between revisions of "1960 AHSME Problems/Problem 20"
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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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Revision as of 18:06, 17 May 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
![\[\binom{8}{3}(\frac{x^2}{2})^{5}(\frac{-2}{x})^3\]](http://latex.artofproblemsolving.com/6/6/6/6662fcbdde87ccdaf30cab5094b7e7f6992c92ac.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 | ||
