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>?
 
  
 
<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 1==
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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>.
 
  
==Solution 2==
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First, note that <math>f(x) = f(1-x)</math>. We can see this since
We can first plug in a few numbers and see what happens. We get <math>(\frac{1}{2019})^2(\frac{2018}{2019})^2 + \frac{2}{2019})^2(\frac{2017}{2019})^2</math> and so on. Then, we can skip to the end and see that the last term and the first term are equal, and cancel each other out because they have different signs. Therefore we see that every number cancels out. It might seem that there is some term in the middle, but if we use a smaller example to check, we see that that is not the case. Therefore, the answer is <math>\boxed{\text{(A) 0}}</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:
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<cmath>\left( f \left(\frac{1}{2019} \right) - f \left(\frac{2018}{2019} \right) \right) +
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\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>
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Now it is clear that all the terms will cancel out (the series telescopes), so the answer is <math>\boxed{\textbf{(A) }0}</math>.
  
-- clara32356 (Claire)
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==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
 

Latest revision as of 18:57, 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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