$\sum\limits_{k=1}^{n^2}E(\sqrt{k})\quad n\in\mathbb{N}^{*}$

Question

I found in my archives solution of this exercise

Calculate $$\sum\limits_{k=1}^{n^2}E(\sqrt{k})\quad n\in\mathbb{N}^{*}$$

$E$ represents the floor function Solution:

they made

Let $S_{n}=\sum\limits_{k=1}^{n^2}E(\sqrt{k})$

we've \begin{align*} S_{n}&=1+\sum_{2}^{3}1+\sum_{4}^{8}2+\sum_{9}^{15}3+\sum_{16}^{24}4+\ldots+\sum_{(n-1)^{2}}^{n^2-1}+n\\ &=1+2.1+5.2+7.3+9.4+\ldots +(2n-1)(n-1)+n\\ &=[\sum_{k=1}^{n}(2k-1)(k-1)]+n\\ &=[\sum_{k=1}^{n}2k^2-3k+1]+n\\ &=2\sum_{k=1}^{n}k^2+-3\sum_{k=1}^{n}k+\sum_{k=1}^{n}1+n \end{align*}

My question i didn't understand how they get this expression and the idea behind it ( is it Change of variables and if its why ): $$S_{n}=1+\sum_{2}^{3}1+\sum_{4}^{8}2+\sum_{9}^{15}3+\sum_{16}^{24}4+\ldots+\sum_{(n-1)^{2}}^{n^2-1}+n$$

I tried to develop the series, but I do not find the same result and what is its role

by the way one of my friend told me $[(j-1)^2;j^2-1]$ but i can see why and how get this idea

how can i go from $1\leq K \leq n^2$ to get that $E(\sqrt{k})=j \iff j^2 \le k < (j+1)^2$ can you elaborate that

any help would be appreciated!

Answer 1

HINT: $\lfloor \sqrt x\rfloor=k$ for $k^2\leq x<(k+1)^2$.

$\def\F#1{\lfloor #1\rfloor}$ It means: \begin{align} &\sqrt{1}+(\F{\sqrt{2}}+\F{\sqrt{3}})+(\F{\sqrt{4}}+\F{\sqrt{5}}+\F{\sqrt{6}}+\F{\sqrt{7}}+\F{\sqrt{8}})+\dots\\ &\quad =1+(1+1)+(2+2+2+2+2)+\dots=\\ &\quad= 1+\sum_{2}^{3}1+\sum_{4}^{8}2+\dots \end{align}

Answer 2

The point is that $E(\sqrt{k}) = j$ iff $j^2 \le k < (j+1)^2$. So there are $(j+1)^2 - j^2$ terms where the summand is $j$.