Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1965 AHSME Problems/Problem 14"

m (typo fix)
m (maa notice)
 
Line 17: Line 17:
 
==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1965|num-b=13|num-a=15}}
 
{{AHSME 40p box|year=1965|num-b=13|num-a=15}}
 +
{{MAA Notice}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 16:01, 18 July 2024

Problem 14

The sum of the numerical coefficients in the complete expansion of $(x^2 - 2xy + y^2)^7$ is:

$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 7 \qquad \textbf{(C) }\ 14 \qquad \textbf{(D) }\ 128 \qquad \textbf{(E) }\ 128^2$

Solution

Notice that the given equation, $(x^2 - 2xy + y^2)^7$ can be factored into $(x-y)^{14} = \sum_{k=0}^{14} \binom{14}{k}(-1)^k\cdot x^{14-k}\cdot y^k$.

Notice that if we plug in $x = y = 1$, then we are simply left with the sum of the coefficients from each term. Therefore, the sum of the coefficients is $(1-1)^7 = 0, \boxed{\textbf{(A)}}$.

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1965_AHSME_Problems/Problem_14&oldid=223024"