Difference between revisions of "1974 AHSME Problems/Problem 3"
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<cmath> (1+2x-x^2)(1+2x-x^2)(1+2x-x^2)(1+2x-x^2) </cmath> | <cmath> (1+2x-x^2)(1+2x-x^2)(1+2x-x^2)(1+2x-x^2) </cmath> | ||
| − | We can now see that the only way to get an <math> x^7 </math> is by taking three <math> -x^2 </math> and one <math> 2x </math>. There are <math> \binom{4}{1}=4 </math> | + | We can now see that the only way to get an <math> x^7 </math> is by taking three <math> -x^2 </math> and one <math> 2x </math>. There are <math> \binom{4}{1}=4 </math> ways to pick which term the <math> 2x </math> comes from, and the coefficient of each one is <math> (-1)^3(2)=-2 </math>. Therefore, the coefficient of <math> x^7 </math> is <math> (4)(-2)=-8, \boxed{\text{A}} </math>. |
==See Also== | ==See Also== | ||
Latest revision as of 19:53, 5 October 2020
Problem
The coefficient of 
in the polynomial expansion of
![\[(1+2x-x^2)^4\]](http://latex.artofproblemsolving.com/2/3/9/2399c6b47d2c9caaf078097a183bf782cfd3abff.png)
is
Solution
Let's write out the multiplication, so that it becomes easier to see.
![\[(1+2x-x^2)(1+2x-x^2)(1+2x-x^2)(1+2x-x^2)\]](http://latex.artofproblemsolving.com/5/7/7/5776793d7b7c952f044d47e440c6b5a144e6a823.png)
We can now see that the only way to get an is by taking three 
and one 
. There are 
ways to pick which term the comes from, and the coefficient of each one is 
. Therefore, the coefficient of is 
.
See Also
| 1974 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
