Problem:
Let f(n) be the integer closest to 4n. Find ∑k=11995f(k)1.
Solution:
Let m be a positive integer. The largest integer n for which f(n)=m is
⌊(m+21)4⌋=⌊m4+2m3+23m2+21m+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∑1995f(k)1=k=1∑1785f(k)1+71995−1785=m=1∑6m4m3+m+30=m=1∑6(4m2+1)+30=400
The problems on this page are the property of the MAA's American Mathematics Competitions