Difference between revisions of "2019 AMC 12B Problems/Problem 8"

Revision as of 17:21, 14 February 2019

Problem

Let $f(x) = x^{2}(1-x)^{2}$ . What is the value of the sum $f(\frac{1}{2019})-f(\frac{2}{2019})+f(\frac{3}{2019})-f(\frac{4}{2019})+\cdots$

$+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})$ ?

$\textbf{(A) }0\qquad\textbf{(B) }\frac{1}{2019^{4}}\qquad\textbf{(C) }\frac{2018^{2}}{2019^{4}}\qquad\textbf{(D) }\frac{2020^{2}}{2019^{4}}\qquad\textbf{(E) }1$

Solution

Note that $f(x) = f(1-x)$ . We can see from this that the terms cancel and the answer is $\boxed{A}$ .

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.

@@ Line 5: / Line 5: @@
 <math>+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})</math>?
-(A) <math>0</math>, (B) <math>\frac{1}{2019^{4}}</math>, (C) <math>\frac{2018^{2}}{2019^{4}}</math>, (D) <math>\frac{2020^{2}}{2019^{4}}</math>, (E) <math>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==

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Difference between revisions of "2019 AMC 12B Problems/Problem 8"

Revision as of 17:21, 14 February 2019

Problem

Solution

See Also