Roots in Floor

We note that $\lfloor \sqrt n \rfloor = k$ for $k^2 \le n \le (k+1)^2 - 1$ , where $n$ and $k$ are positive integers, and we have $(k+1)^2 - k^2 = 2k+1$ such $n$ . Therefore the sum:

$\begin{aligned} \sum_{n=1}^{145} \left \lfloor \sqrt n \right \rfloor & = \sum_{n=1}^{12^2-1} \left \lfloor \sqrt n \right \rfloor + \sum_{n=144}^{145} \left \lfloor \sqrt n \right \rfloor \\ & = \sum_{k=1}^{11} k(2k+1) + \left \lfloor \sqrt{144} \right \rfloor + \left \lfloor \sqrt{145} \right \rfloor \\ & = \sum_{k=1}^{11} (2k^2+k) + 12 + 12 \\ & = 2\cdot \frac {11(12)(23)}6 + \frac {11(12)}2 + 24 \\ & = 1012 + 66 + 24 \\ & = \boxed{1102} \end{aligned}$