Difference between revisions of "2019 AMC 12B Problems/Problem 8"
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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 | ||
| − | < | + | <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>\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> | ||
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First, note that <math>f(x) = f(1-x)</math>. We can see this since | 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> | + | <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> |
| − | + | Using this result, we regroup the terms accordingly: | |
<cmath>\left( f \left(\frac{1}{2019} \right) - f \left(\frac{2018}{2019} \right) \right) + | <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 | \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2017}{2019} \right) \right) + \cdots | ||
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\left( f \left(\frac{2}{2019} \right) - f \left(\frac{2}{2019} \right) \right) + \cdots | \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> | + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1009}{2019} \right) \right)</cmath> | ||
| − | Now | + | Now it is clear that all the terms will cancel out (the series telescopes), so the answer is <math>\boxed{\textbf{(A) }0}</math>. |
| + | |||
| + | == Video Solution == | ||
| + | https://youtu.be/ba6w1OhXqOQ?t=2457 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=4076 | ||
| + | |||
| + | - AMBRIGGS | ||
==See Also== | ==See Also== | ||
Revision as of 13:14, 30 July 2022
Contents
Problem
Let 
. What is the value of the sum
![\[f \left(\frac{1}{2019} \right)-f \left(\frac{2}{2019} \right)+f \left(\frac{3}{2019} \right)-f \left(\frac{4}{2019} \right)+\cdots + f \left(\frac{2017}{2019} \right) - f \left(\frac{2018}{2019} \right)?\]](http://latex.artofproblemsolving.com/b/b/c/bbc10c8e158e346ff887ab29152c7d1b5a7a5922.png)
Solution
First, note that 
. We can see this since
![\[f(x) = x^2(1-x)^2 = (1-x)^2x^2 = (1-x)^{2}\left(1-\left(1-x\right)\right)^{2} = f(1-x)\]](http://latex.artofproblemsolving.com/2/1/0/210ab2d10107fd29d82e3953bc19115911cd7dce.png)
Using this result, we regroup the terms accordingly:
![\[\left( f \left(\frac{1}{2019} \right) - f \left(\frac{2018}{2019} \right) \right) + \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2017}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1010}{2019} \right) \right)\]](http://latex.artofproblemsolving.com/b/4/9/b4920afe33917c882c3ea0d6b0594effda45801e.png)
![\[= \left( f \left(\frac{1}{2019} \right) - f \left(\frac{1}{2019} \right) \right) + \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1009}{2019} \right) \right)\]](http://latex.artofproblemsolving.com/8/7/3/87361380209081cfc0c5667d1224d7c7ce591457.png)
Now it is clear that all the terms will cancel out (the series telescopes), so the answer is 
.
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=2457
~ pi_is_3.14
https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=4076
- AMBRIGGS
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. 
