Floor square sum

We can notice that,

$\begin{aligned} \color{#3D99F6}{m} \text{ values } \begin{cases} \lfloor \sqrt{n^2} \rfloor & =n \\ \lfloor \sqrt{n^2+1} \rfloor & =n \\ \quad \quad \ldots \\ \lfloor \sqrt{(n+1)^2-1}\rfloor & =n \end{cases} \\ \color{#3D99F6}{m}=(n+1)^2-1-n^2+1 & =2n+1 \end{aligned}$

And also see that, $44^2=1936<1988<2025=45^2$

Therefore, $S$ can be written as,

$\begin{aligned} S & =\displaystyle \sum_{1≤n≤43}n(2n+1)+44(53) \\ & = \displaystyle \sum_{1≤n≤43}(2n^2+n)+2332 \\ & = 2 \displaystyle \sum_{1≤n≤43}n^2+ \displaystyle \sum_{1≤n≤43}n+2332 \\ & = 2\cdot \frac{(43)(44)(87)}{6}+\frac{(43)(44)}{2}+2332 \\ & =\boxed{58146}\end{aligned}$

$\lfloor\sqrt{1988}\rfloor=44 \\ 44^2=1936 \\1988-1935=53\\ 53 \times 44 = 2332\\ \text{There are }2n+1\text{ numbers that takes the value n. So}\\ S=2332+\sum\limits_{n=1}^{43}{n(2n+1)}\\ S=2332+\sum\limits_{n=1}^{43}{(2n^2+n)}\\ S= 2332+\sum\limits_{n=1}^{43}{2n^2}+\sum\limits_{n=1}^{43}{n}\\ S=2332+\frac{43 \times 44 \times 87}{3}+\frac{43 \times 44}{2}\\ S =2332+54868+946\\ S={\boxed{58146}}$

The series, S , shows an apparent pattern. In approaching the problem, what must be done first is to find the greatest perfect square integer $<1988$ . $1988$ is an integer $\in$ $[44^{2},45^{2}]$ , so it is obvious that we are to choose $44^{2}$ because $1936<1988$ . Now that we have chosen our upper bound of summation, we proceed to the establishment of our sum. So we would have $(44 \times 53)+\sum_{i=1}^{43} (n(2n+1))$ $=$ $2332+\sum_{i=1}^{43} (2n^{2}+n)$ $=$ $2332+2 \times \sum_{i=1}^{43} n^{2}+\sum_{i=1}^{43} n$ . At this point, we will use the formula for the sum of the squares of the first $n$ natural numbers and for the sum of the first $n$ natural numbers: $S_{n_{squared}}=\frac{[n(n+1)(2n+1)]}{6}; S_{n}=\frac{[n(n+1)]}{2}$ , respectively. By plugging in $n=43$ , we yield $2332+2[\frac{(43)(43+1)(2(43)+1)}{6}]+\frac{(43)(43+1)}{2}$ which simplifies to $2332+(43 \times 44 \times 29)+(43 \times 22)$ $=$ $58146$ . $\therefore S=\lfloor\sqrt{1}\rfloor+\lfloor\sqrt{2}\rfloor+\lfloor\sqrt{3}\rfloor+...+\lfloor\sqrt{1988}\rfloor=\boxed{58146}$