2019 AMC 12B Problems/Problem 8: Difference between revisions

Revision as of 12:37, 14 February 2019

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

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

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

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

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 ==Solution==
-Note that <math>f(x) = f(1-x)</math>. We can see from this that the terms cancel and the answer is <math>\boxed{A}</math>. -zachc16
+Note that <math>f(x) = f(1-x)</math>. We can see from this that the terms cancel and the answer is <math>\boxed{A}</math>.
 ==See Also==
 {{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}}
 {{MAA Notice}}