# Difference between revisions of "2019 AMC 12B Problems/Problem 8"

## 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 $\boxed{\text{(A) 0}}$.