Problem:
Let f(x)=x2(1−x)2. What is the value of the sum
f(20191​)−f(20192​)+f(20193​)−(20194​)+⋯+f(20192017​)−f(20192018​)?​
Answer Choices:
A. 0
B. 201941​
C. 2019420182​
D. 2019420202​
E. 1
Solution:
First, note that f(x)=f(1−x). We can see this since
f(x)=x2(1−x)2=(1−x)2x2=(1−x)2(1−(1−x))2=f(1−x)
Using this result, we regroup the terms accordingly:
(f(20191​)−f(20192018​))+(f(20192​)−f(20192017​))+⋯+(f(20191009​)−f(20191010​))=(f(20191​)−f(20191​))+(f(20192​)−f(20192​))+⋯+(f(20191009​)−f(20191009​))
Now it is clear that all the terms will cancel out (the series telescopes), so the answer is (A)0​ .
The problems on this page are the property of the MAA's American Mathematics Competitions