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