Difference between revisions of "2019 AMC 12B Problems/Problem 8"
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==Problem== | ==Problem== | ||
Let <math>f(x) = x^{2}(1-x)^{2}</math>. What is the value of the sum | Let <math>f(x) = x^{2}(1-x)^{2}</math>. What is the value of the sum | ||
− | <math>f(\frac{1}{2019})-f(\frac{2}{2019})+f(\frac{3}{2019})-f(\frac{4}{2019})+\cdots </math> | + | <math>f\left(\frac{1}{2019} \right)-f \left(\frac{2}{2019} \right)+f \left(\frac{3}{2019} \right)-f \left(\frac{4}{2019} \right)+\cdots </math> |
− | <math>+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})</math>? | + | <math>+ f \left(\frac{2017}{2019} \right) - f \left(\frac{2018}{2019} \right)</math>? |
<math>\textbf{(A) }0\qquad\textbf{(B) }\frac{1}{2019^{4}}\qquad\textbf{(C) }\frac{2018^{2}}{2019^{4}}\qquad\textbf{(D) }\frac{2020^{2}}{2019^{4}}\qquad\textbf{(E) }1</math> | <math>\textbf{(A) }0\qquad\textbf{(B) }\frac{1}{2019^{4}}\qquad\textbf{(C) }\frac{2018^{2}}{2019^{4}}\qquad\textbf{(D) }\frac{2020^{2}}{2019^{4}}\qquad\textbf{(E) }1</math> | ||
− | ==Solution | + | ==Solution== |
− | + | First, note that <math>f(x) = f(1-x)</math>. We can see this since | |
− | + | <cmath>f(x) = x^2(1-x)^2 = (1-x)^2x^2 = f(1-x)</cmath> | |
− | + | From this, we regroup the terms accordingly: | |
− | + | <cmath>\left( f \left(\frac{1}{2019} \right) - f \left(\frac{2018}{2019} \right) \right) + | |
− | + | \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2017}{2019} \right) \right) + \cdots | |
− | + | + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1010}{2019} \right) \right)</cmath> | |
− | + | <cmath> = \left( f \left(\frac{1}{2019} \right) - f \left(\frac{1}{2019} \right) \right) + | |
+ | \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2}{2019} \right) \right) + \cdots | ||
+ | + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1009}{2019} \right) \right)</cmath> | ||
+ | Now, it is clear that all the terms will cancel out, and so the answer is <math>\boxed{\text{(A) 0}}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}} | {{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
SUB2PEWDS | SUB2PEWDS |
Revision as of 22:34, 16 February 2019
Problem
Let . What is the value of the sum
?
Solution
First, note that . We can see this since From this, we regroup the terms accordingly: Now, it is clear that all the terms will cancel out, and so the answer is .
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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