2019 AMC 12B Problems/Problem 8

Revision as of 22:34, 16 February 2019 by Vedadehhc (talk | contribs) (combined and reworded solutions for clarity and brevity)

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 (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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