Difference between revisions of "2019 AMC 12B Problems/Problem 8"
Sevenoptimus (talk | contribs) m (Fixed the problem statement) |
|||
(4 intermediate revisions by 2 users not shown) | |||
Line 1: | Line 1: | ||
==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 | ||
− | < | + | <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> | ||
− | ==Solution | + | ==Solution== |
− | |||
− | == | + | 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 = (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) + | ||
+ | \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)</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 (the series telescopes), so the answer is <math>\boxed{\textbf{(A) }0}</math>. | ||
− | + | ==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}} | ||
− |
Latest revision as of 18:57, 18 February 2019
Problem
Let . What is the value of the sum
Solution
First, note that . We can see this since Using this result, we regroup the terms accordingly: Now it is clear that all the terms will cancel out (the series telescopes), so the answer is .
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.