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

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{{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}}
{{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}}
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{{MAA Notice}}

Revision as of 17:05, 15 February 2019


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

$+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})$?

$\textbf{(A) }0\qquad\textbf{(B) }\frac{1}{2019^{4}}\qquad\textbf{(C) }\frac{2018^{2}}{2019^{4}}\qquad\textbf{(D) }\frac{2020^{2}}{2019^{4}}\qquad\textbf{(E) }1$

Solution 1

Note that $f(x) = f(1-x)$. We can see from this that the terms cancel and the answer is $\boxed{A}$.

Solution 2

We can first plug in a few numbers and see what happens. We get $(\frac{1}{2019})^2(\frac{2018}{2019})^2 + \frac{2}{2019})^2(\frac{2017}{2019})^2$ and so on. Then, we can skip to the end and see that the last term and the first term are equal, and cancel each other out because they have different signs. Therefore we see that every number cancels out. It might seem that there is some term in the middle, but if we use a smaller example to check, we see that that is not the case. Therefore, the answer is $\boxed{\text{(A) 0}}$

-- clara32356 (Claire)

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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