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

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==Problem==
 
==Problem==
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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>
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<cmath>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 + f \left(\frac{2017}{2019} \right) - f \left(\frac{2018}{2019} \right)?</cmath>
  
<math>+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})</math>?
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<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>
  
(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>.
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==Solution==
  
==Solution==
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First, note that <math>f(x) = f(1-x)</math>. We can see this since
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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<cmath>f(x) = x^2(1-x)^2 = (1-x)^2x^2 = (1-x)^{2}\left(1-\left(1-x\right)\right)^{2} = f(1-x)</cmath>
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Using this result, 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
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+ \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1010}{2019} \right) \right)</cmath>
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<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
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+ \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 (the series telescopes), so the answer is <math>\boxed{\textbf{(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}}

Revision as of 19:57, 18 February 2019

Problem

Let $f(x) = x^{2}(1-x)^{2}$. What is the value of the sum \[f \left(\frac{1}{2019} \right)-f  \left(\frac{2}{2019} \right)+f \left(\frac{3}{2019} \right)-f \left(\frac{4}{2019} \right)+\cdots + f \left(\frac{2017}{2019} \right) - f \left(\frac{2018}{2019} \right)?\]

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

Solution

First, note that $f(x) = f(1-x)$. We can see this since \[f(x) = x^2(1-x)^2 = (1-x)^2x^2 = (1-x)^{2}\left(1-\left(1-x\right)\right)^{2} = f(1-x)\] Using this result, we regroup the terms accordingly: \[\left( f \left(\frac{1}{2019} \right) - f \left(\frac{2018}{2019} \right) \right) +  \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2017}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1010}{2019} \right) \right)\] \[= \left( f \left(\frac{1}{2019} \right) - f \left(\frac{1}{2019} \right) \right) +  \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1009}{2019} \right) \right)\] Now it is clear that all the terms will cancel out (the series telescopes), so the answer is $\boxed{\textbf{(A) }0}$.

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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