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

(Created page with "== Problem 14== The sum of the numerical coefficients in the complete expansion of <math>(x^2 - 2xy + y^2)^7</math> is: <math>\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 7 \qqua...")
 
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Notice that the given equation, <math>(x^2 - 2xy + y^2)^7</math> can be factored into <math>(x-y)^{14} = \sum_{k=0}^{14} \binom{14}{k}(-1)^k\cdot x^{14-k}\cdot y^k</math>.  
 
Notice that the given equation, <math>(x^2 - 2xy + y^2)^7</math> can be factored into <math>(x-y)^{14} = \sum_{k=0}^{14} \binom{14}{k}(-1)^k\cdot x^{14-k}\cdot y^k</math>.  
  
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Notice that if we plug in <math>x = y</math> = 1, then we are simply left with the sum of the coefficients from each term. Therefore, the sum of the coefficients is <math>(1-1)^2 = 0, \boxed{\textbf{(A)}}</math>
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Notice that if we plug in <math>x = y = 1</math>, then we are simply left with the sum of the coefficients from each term. Therefore, the sum of the coefficients is <math>(1-1)^7 = 0, \boxed{\textbf{(A)}}</math>.
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==See Also==
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{{AHSME 40p box|year=1965|num-b=13|num-a=15}}
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{{MAA Notice}}
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[[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

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