Find $[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+....+[\sqrt{2019}]$ 
Let [$x$] denote the greatest integer less than or equal to $x$. Find the value of $$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+....+[\sqrt{2019}]$$

I was able to find out the general pattern in the series i.e. $3(1)+ 5(2)+ 7(3)+\ldots+87(43)$ but was unable to proceed further.
 A: Hint 1:   $[\sqrt{n^2}], [\sqrt{n^2 +1}], .........[\sqrt{(n+1)^2 - 1}]$ all equal the same the same thing: $n$.
Hint 2: $(n+1)^2 - 1= n^2 + 2n$ so from $[\sqrt{n^2}]$ to $[\sqrt{(n+1)^2 - 1}]=[\sqrt{n^2 + 2n}]$ there are $2n+1$ terms.
Hint 3: So  we have $\color{blue}{[\sqrt {1}] + [\sqrt 2]+ [\sqrt 3]} + \color{purple}{[\sqrt 4] + ....+[\sqrt 8]} +\color{orange}{[\sqrt 9]+ ....+[\sqrt{15}]}.....+\color{green}{[\sqrt M^2]+...... + [\sqrt {(M+1)^2-1}]} + \color{red}{[\sqrt {(M+1)^2}] + ...... + \sqrt{2019}}$  where $(M+1)^2 \le 2019 < (M+2)^2$.
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Solution:
$M+1= 44$ and $44^2 = 1936$ so so we have
$\color{blue}{1 + 1+ 1} + \color{purple}{2 + ....+2} +\color{orange}{3+ ....+3}.....+\color{green}{43+...... + 43} + \color{red}{44 + ...... + 44}$
$\color{blue}{1\cdot (2\cdot 1 + 1)} + \color{purple}{2\cdot (2\cdot 2 + 1)} +\color{orange}{3\cdot (2\cdot 3 + 1)}.....+\color{green}{43\cdot (2\cdot 43 + 1)} + \color{red}{44\times(2019-1936+1)}=$
$\sum_{k=1}^{43}[k(2k+1)] + 44\cdot 84=$
$2\sum_{k=1}^{43} k^2+\sum_{k=1}^{43} k + 3696$
..... and that's it.  Apply $\sum_{k=1}^n k = \frac {n(n+1)}2$ and $\sum_{k=1}^n k^2 = \frac {n(n+1)(2n+1)}6$ and you are done.
