Long computation

Let $m = \left \lceil \sqrt{k} \right \rceil$ and $n = \left \lfloor \sqrt{m} \right \rfloor$ . Then, we have:

\(\begin{array} {} n = \left \lfloor \sqrt{m} \right \rfloor & \Rightarrow n^2 \le m \le (n+1)^2 - 1 \\ m = \left \lceil \sqrt{k} \right \rceil & \Rightarrow (m-1)^2+1 \le k \le m^2 & \Rightarrow k \le \left[(n+1)^2 - 1\right]^2 \end{array} \)

Therefore,

$\begin{array} {llc} n = 1 & \Rightarrow & 1 \le k \le 9 \\ n = 2 & \Rightarrow & 10 \le k \le 64 \\ n = 3 & \Rightarrow & 65 \le k \le 225 \\ n = 4 & \Rightarrow & 226 \le k \le 576 \\ n = 5 & \Rightarrow & 577 \le k \le 578 \end{array}$

And,

$\begin{aligned} \sum_{k=1}^{578} \left \lfloor \sqrt{\left \lceil \sqrt{k} \right \rceil} \right \rfloor & = \sum_{k=1}^{9} 1 + \sum_{k=10}^{64} 2 + \sum_{k=65}^{225} 3 + \sum_{k=226}^{576} 4 + \sum_{k=577}^{578} 5 \\ & = 9(1) + 55(2) + 161(3) + 351(4) + 2(5) \\ & = 9 + 110 + 483 + 1404 + 10 \\ & = \boxed{2016} \end{aligned}$

First, notice that $\lfloor{\sqrt{k}}\rfloor=n$ when $n^2\leq k<(n+1)^2$ .

Also notice that $\lceil{\sqrt{k}}\rceil=n$ when $(n-1)^2<k\leq n^2$ . If we want to find when $n^2\leq\lceil{\sqrt{k}}\rceil<(n+1)^2$ (this is due to the bounds we found above), simply take the lower bound of when $\lceil{\sqrt{k}}\rceil=n^2$ and the upper bound of $\lceil{\sqrt{k}}\rceil=(n+1)^2$ . This gives us the following result: $\lfloor{\sqrt{\lceil{\sqrt{k}}\rceil}}\rfloor=n\text{ when }(n^2-1)^2<k\leq ((n+1)^2-1)^2$

To find how many numbers are in that bound, simply subtract the lower bound from the upper bound:

$((n+1)^2-1)^2-(n^2-1)^2=4n^3+6n^2-1$

Note that $576=((4+1)^2-1)^2$ , so we can take our sum from $1$ to $576$ and then there are 2 terms left over (both will result in 5). Taking the sum from $1$ to $4$ and counting our two that our formula misses:

$5+5+\displaystyle\sum_{n=1}^4 n(4n^3+6n^2-1)=5+5+2006=\boxed{2016}$

I did it the long way

Wrote a c++ program to find it

#include <iostream>
#include<math.h>
#include <cstdlib>
using namespace std;
float fun (int k){
return (sqrt(sqrt(k)));

}
int main() {
        float i,sum=0;
        for (i=1;i<579;i++){
        sum+=fun(i);
    }
    cout<<sum;
    return 0;
}

but it outputs 2269.38 if i take only integers i get 1962 why is it that we get different answers from computer and actual solution

Did the long way but still got the same answer