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== | ||
{{AHSME box|year=1974|num-b=2|num-a=4}} | {{AHSME box|year=1974|num-b=2|num-a=4}} | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:53, 5 October 2020
Problem
The coefficient of in the polynomial expansion of
is
Solution
Let's write out the multiplication, so that it becomes easier to see.
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 | |
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All AHSME Problems and Solutions |
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