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Difference between revisions of "2019 AMC 12B Problems/Problem 8"
(combined and reworded solutions for clarity and brevity)
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==Problem==
 
==Problem==
 
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>
+
<math>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 </math>
  
−
<math>+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})</math>?
+
<math>+ f \left(\frac{2017}{2019} \right) - f \left(\frac{2018}{2019} \right)</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==
+
==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>.
+
First, note that <math>f(x) = f(1-x)</math>. We can see this since
−
 
+
<cmath>f(x) = x^2(1-x)^2 = (1-x)^2x^2 = f(1-x)</cmath>
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==Solution 2==
+
From this, we regroup the terms accordingly:
−
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>
+
<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
−
-- clara32356 (Claire)
+
+ \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1010}{2019} \right) \right)</cmath>
−
 
+
<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
 +
+ \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, and so the answer is <math>\boxed{\text{(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}}
 
SUB2PEWDS
 
SUB2PEWDS

Revision as of 23:34, 16 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 \[f(x) = x^2(1-x)^2 = (1-x)^2x^2 = f(1-x)\] From this, 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, and so the answer is $\boxed{\text{(A) 0}}$.

See Also

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. AMC logo.png

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