Problem:
Find the integer that is closest to 1000∑n=310000​n2−41​.
Solution:
Because n2−41​=41​(n−21​−n+21​), the series telescopes, and it follows that
1000n=3∑10000​n2−41​​=250(11​+21​+31​+41​−99991​−100001​−100011​−100021​)=250+125+3250​+4250​−9999250​−10000250​−10001250​−10002250​=520+65​−r​
where the positive number r is less than 1/3. Thus the requested integer is 521​.
The problems on this page are the property of the MAA's American Mathematics Competitions