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

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==Solution==
 
==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>
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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
  
 
==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}}

Revision as of 12:36, 14 February 2019

Problem

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

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

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

Solution

Note that $f(x) = f(1-x)$. We can see from this that the terms cancel and the answer is $\boxed{A}$. -zachc16

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
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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