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

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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>
==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}}
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SUB2PEWDS

Revision as of 22:34, 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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