Difference between revisions of "1965 AHSME Problems/Problem 14"
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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>. | ||
| − | 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)^ | + | 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>. |
==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 
is:
Solution
Notice that the given equation, can be factored into 
.
Notice that if we plug in 
, then we are simply left with the sum of the coefficients from each term. Therefore, the sum of the coefficients is 
.
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. 
