AMC12 2019 B

AMC12 2019 B · Q8

AMC12 2019 B · Q8. It mainly tests Functions basics, Basic counting (rules of product/sum).

Let $f(x) = x^2(1-x)^2$. What is the value of the sum $$f\left(\frac{1}{2019}\right) - f\left(\frac{2}{2019}\right) + f\left(\frac{3}{2019}\right) - f\left(\frac{4}{2019}\right) + \dots + f\left(\frac{2017}{2019}\right) - f\left(\frac{2018}{2019}\right)$$

设 $f(x) = x^2(1-x)^2$。下列和的值是多少 $$f\left(\frac{1}{2019}\right) - f\left(\frac{2}{2019}\right) + f\left(\frac{3}{2019}\right) - f\left(\frac{4}{2019}\right) + \dots + f\left(\frac{2017}{2019}\right) - f\left(\frac{2018}{2019}\right)$$

(A) 0 0

(B) \frac{1}{2019^4} \frac{1}{2019^4}

(D) \frac{2020^2}{2019^4} \frac{2020^2}{2019^4}

(E) 1 1

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Note that $f(1-x)=(1-x)^2(1-(1-x))^2=(1-x)^2x^2=f(x).$ Therefore $f\left(\frac{1}{2019}\right)=f\left(\frac{2018}{2019}\right)$, so the first and last terms of the given sum combine to give 0. Similarly, $f\left(\frac{2}{2019}\right)-f\left(\frac{2017}{2019}\right)=0$, $f\left(\frac{3}{2019}\right)-f\left(\frac{2016}{2019}\right)=0$, and so on. Because there are an even number of terms, all the terms cancel out in pairs, and the sum is 0.

答案（A）：注意到 $f(1-x)=(1-x)^2(1-(1-x))^2=(1-x)^2x^2=f(x).$ 因此 $f\left(\frac{1}{2019}\right)=f\left(\frac{2018}{2019}\right)$，所以所给求和式的第一项与最后一项相加为 0。类似地，$f\left(\frac{2}{2019}\right)-f\left(\frac{2017}{2019}\right)=0$，$f\left(\frac{3}{2019}\right)-f\left(\frac{2016}{2019}\right)=0$，依此类推。由于项数为偶数，所有项都两两相消，因此和为 0。

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