How to solve $\sum_{k=1}^{2500}\left \lfloor{\sqrt{k}}\right \rfloor $? I was trying to solve $\sum_{k=1}^{2500}\left \lfloor{\sqrt{k}}\right \rfloor $ using  Iverson's brackets but I can't get the bounds right. I think I'm also missing something.
Here's what I did:
$ m = \left \lfloor{\sqrt{k}}\right \rfloor$
$\sum_{m,k} \space m \space[ m = \left \lfloor{\sqrt{k}}\right \rfloor][0 \leq k < 2,500]$
$\sum_{m,k} \space m \space [ m^2 \leq k < (m+1)^2 ][0 \leq k \leq 2,500]$
$\sum_{m,k} \space m \space[ m^2 \leq k \leq 2,500 < (m+1)^2 ]$
from here I'm not sure what will be the bounds of $m$. I'm new to this kind of sum manipulation so please bear with me.
 A: You have
$$\sum_{m=1}^{50}m\sum_k[m^2\le k<\min\left((m+1)^2,2501\right)].$$
The inner sum is just the number of $k$ for which the inequality holds.
For $m<50$ that number is just $(m+1)^2-m^2$, and for $m=50$ it's $1$. So
one gets
$$\sum_{m=1}^{49}m\left((m+1)^2-m^2\right)+50.$$
A: Hint: You know that $\lfloor\sqrt{k}\rfloor = m$ if and only if $m^2 \le k \le (m+1)^2-1$. Hence:
$\lfloor\sqrt{k}\rfloor = 1$ for $1 \le k \le 3$ ($3$ values of $k$)
$\lfloor\sqrt{k}\rfloor = 2$ for $4 \le k \le 8$ ($5$ values of $k$)
$\lfloor\sqrt{k}\rfloor = 3$ for $9 \le k \le 15$ ($7$ values of $k$)
...
$\lfloor\sqrt{k}\rfloor = 49$ for $49^2 \le k \le 50^2-1 = 2499$ ($?$ values of $k$)
$\lfloor\sqrt{k}\rfloor = 50$ for $k = 2500 = 50^2$ ($1$ value of $k$)
Can you use this information to evaluate the sum?
A: We have $\lfloor \sqrt{k}\rfloor=m\,\, m^2\le k<(m+1)^2$. Thus, $$\sum_{k=1}^{2500}\lfloor \sqrt{k}\rfloor$$$$=1+1+1_{3\, \text{times}}+2+\ldots+2_{5\,\text{times}}+\ldots+k+\ldots+k_{2k+1\,\text{times}}+49+49+\ldots+49_{99\,\text{times}}+50$$$$=\left[\sum_{k=1}^{49}k(2k+1)\right]+50$$$$=\left[\sum_{k=1}^{49}2k^2+k\right]+50$$$$=\frac{49(50)(99)}{3}+50$$$$=80850+50=80900.$$
