2019 AMC 12B Problem 8

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

f(x)=f(1-x)

Pair

f\!\left(\tfrac{k}{2019}\right)

with

f\!\left(\tfrac{2019-k}{2019}\right),

and compare their signs

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.

Problem 8 in other years