Let f(x)=x^2(1-x)^2. What is the value of the sum f ((1)/(2019))-f ((2)/(2019))+f ((3)/(2019)) -f ((4)/(2019))+\cdots+f ((2017)/(2019))-f ((2018)/(2019))?

2019 AMC 12B Problem 8 — Solution

2019 AMC 12B Problem 8

Below is the professionally curated solution for Problem 8 of the 2019 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 12B solutions, or check the answer key.

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Concepts:symmetry (algebra)pairing and grouping

Difficulty rating: 1560

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

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

$0$

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

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

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

$1$

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.

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