Question

Let $f(x)=x^2(1-x)^2.$ What is the value of the sum

$\begin{gathered} f\!\left(\tfrac{1}{2019}\right)-f\!\left(\tfrac{2}{2019}\right) \\ {}+f\!\left(\tfrac{3}{2019}\right)-f\!\left(\tfrac{4}{2019}\right) \\ {}+\cdots+f\!\left(\tfrac{2017}{2019}\right) \\ {}-f\!\left(\tfrac{2018}{2019}\right)? \end{gathered}$

$0$

$\dfrac{1}{2019^4}$

$\dfrac{2018^2}{2019^4}$

$\dfrac{2020^2}{2019^4}$

$1$

Answer: A

Concepts:symmetry (algebra)pairing and grouping

Difficulty rating: 1560

Solution:

Since $f(1-x)=(1-x)^2x^2=f(x),$ we have $f\!\left(\tfrac{k}{2019}\right)=f\!\left(\tfrac{2019-k}{2019}\right).$

In the sum, the term with index $k$ has sign $(-1)^{k+1},$ while the term with index $2019-k$ equals it in value but has sign $(-1)^{2019-k+1}=(-1)^{k},$ the opposite.

Every term cancels with its partner, so the total is $0.$

Thus, A is the correct answer.

2019 AMC 12B Problem 8

Answer: A

Difficulty rating: 1560

Solution:

Problem 8 in Other Years