Question 1209977
<pre>
For a positive integer n, let f(n) denote the integer that is closest to
{{{root(4,n)}}}.  Find the integer m so that
{{{sum( f(n), n=1, m )}}}{{{""=""}}}{{{100}}}.


{{{0.5<root(4,n)<1.5}}}
{{{0.635<n<5.0625}}}
So terms 1 through 5 are all 1's.  That's 5-1+1 = 5 terms of 1's
We have 5 terms so far. And so far the sum is 5x1 = 5.


{{{1.5<root(4,n)<2.5}}}
{{{5.0625<n<39.0625}}}
So terms 6 through 39 are all 2's.  That's 39-6+1=34 terms of 2's.
We have 5+34=39 terms so far. And so far the sum is 5 + 34x2 = 5 + 68 = 73


{{{2.5<root(4,n)<3.5}}}
{{{39.0625<n<150.0625}}}
So terms from 39 on are all 3's.  So we only need to know how many more terms of
3's we need beyond the 39 terms to raise the sum of 73 to 100.

We need to raise the sum by 100-73=27 which will require 27/3=9 more terms of 3's.

So the total number of terms will be 39+9 = 48.

So m = 48

Edwin</pre>