Problem:
Let f(n) be the integer closest to 4n​. Find ∑k=11995​f(k)1​.
Solution:
Let m be a positive integer. The largest integer n for which f(n)=m is
⌊(m+21​)4⌋​=⌊m4+2m3+23​m2+21​m+161​⌋=⌊m4+2m3+21​(3m2+m)+161​⌋=(m+21​)4−161​​
the last line following because 3m2+m is even. Therefore the number of integers n with f(n)=m is
⌊(m+21​)4⌋−⌊((m−1)+21​)4⌋=(m+21​)4−(m−21​)4=4m3+m
Thus f(n)=m for 4m3+m consecutive positive integers n. Now observe that 64<1995<74, so that either f(1995)=6 or f(1995)=7. Because
m=1∑6​(4m3+m)=1785
it follows that f(1786)=f(1787)=⋯=f(1995)=7, hence that
k=1∑1995​f(k)1​​=k=1∑1785​f(k)1​+71995−1785​=m=1∑6​m4m3+m​+30=m=1∑6​(4m2+1)+30=400​​
The problems on this page are the property of the MAA's American Mathematics Competitions