Problem:
Find the integer that is closest to 1000∑n=310000n2−41.
Solution:
Because n2−41=41(n−21−n+21), the series telescopes, and it follows that
1000n=3∑10000n2−41=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