Let $\left[\alpha \right]$ denote the greatest integer $\le \alpha$. Then $\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]+.............+\left[\sqrt{120}\right]$ is equal to

Given,

 $\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]+.............+\left[\sqrt{120}\right]$

Now we now that,

$\left[\sqrt{x}\right]=1, \text{when} x\in [1,4)$, $\left[\sqrt{x}\right]=2, \text{when} x\in [4,9)$ and similarly $\left[\sqrt{x}\right]=10, \text{when} x\in [100,120)$

Now using the above formula we get,

$$\begin{gathered} E=1+1+1+2+2+2+2+2+3+3+3+3+3 \\ +3+3+4+4+\dots \end{gathered}$$

$\Rightarrow E=3\times 1+5\times 2+7\times 3+\dots ..+19\times 9+10\times 21$

$\Rightarrow E=\sum _{r=1}^{10}\left(2r+1\right)r$

$\Rightarrow E=2\sum _{r=1}^{10}\left(r^2+r\right)$

$\Rightarrow E=2\left[\frac{10\times 11\times 21}{6}\right]+\frac{10\times 11}{2}$

$\Rightarrow E=770+55$

$\Rightarrow E=825$