Sum : $\sum \sin \left( \frac{(2\lfloor \sqrt{kn} \rfloor +1)\pi}{2n} \right)$. Calculate : $$ \sum_{k=1}^{n-1} \sin \left( \frac{(2\lfloor \sqrt{kn} \rfloor +1)\pi}{2n} \right).$$
 A: Lemma Summation by Pasts (1)
$$\sum_{k=a}^b f_k\Delta g_k=f_kg_k\Bigg|_{k=a}^{b+1}-\sum_{k=a}^b g_{k+1}\Delta f_k=f_{b+1}g_{b+1}-f_ag_a-\sum_{k=a}^b g_{k+1}\Delta f_k$$
$\displaystyle \text{with difference operator }\Delta :\qquad \Delta f_k=f_{k+1}-f_k$
My solution
$\text{Get }j=\left\lfloor \sqrt{kn}\right\rfloor\Rightarrow j\le \sqrt{kn}<j+1\Rightarrow \dfrac{j^2}{n}\le k<\dfrac{(j+1)^2}{n}$
Therefore if $\left\lfloor\dfrac{j^2}{n}\right\rfloor+1\le k\le \left\lfloor\dfrac{(j+1)^2}{n}\right\rfloor$ then $\sqrt{kn}=j$
Note:
$j=0\Rightarrow \left\lfloor\dfrac{j^2}{n}\right\rfloor+1=1$
$j=n-2\Rightarrow \left\lfloor\dfrac{(j+1)^2}{n}\right\rfloor=n-2$
$j=n-1\Rightarrow \left\lfloor\dfrac{(j+1)^2}{n}\right\rfloor=n>n-1$
So that sum became 
\begin{align*}S&=\sum_{k=1}^{n-1}\sin\left(\dfrac{\left(2\lfloor\sqrt{kn}\rfloor+1\right)\pi}{2n}\right)\\ &=\sin\left(\dfrac{\left(2\lfloor\sqrt{(n-1)n}\rfloor+1\right)\pi}{2n}\right)+\sum_{k=1}^{n-2}\sin\left(\dfrac{\left(2\lfloor\sqrt{kn}\rfloor+1\right)\pi}{2n}\right)\\ &=\sin\left(\frac{(2n-1)\pi}{2n}\right)+\sum_{j=0}^{n-2}\left(\left\lfloor\frac{(j+1)^2}{n}\right\rfloor-\left\lfloor\frac{j^2}{n}\right\rfloor\right)\sin\left(\frac{(2j+1)\pi}{2n}\right)\\ &=\sin\left(\frac{\pi}{2n}\right)+\sum_{j=0}^{n-2}\Delta\left(\left\lfloor\frac{j^2}{n}\right\rfloor\right)\sin\left(\frac{(2j+1)\pi}{2n}\right)\end{align*}
Using the 1, we get
\begin{align*}S&=\sin\left(\frac{\pi}{2n}\right)+\left[\sin\left(\frac{(2j+1)\pi}{2n}\right)\left\lfloor\frac{j^2}{n}\right\rfloor\right]_{j=0}^{n-1}\\ &{}\quad -\sum_{j=0}^{n-2}\left\lfloor\frac{(j+1)^2}{n}\right\rfloor\left(\sin\left(\frac{(2j+3)\pi}{2n}\right)-\sin\left(\frac{(2j+1)\pi}{2n}\right)\right)\\ &=(n-1)\sin\left(\frac{\pi}{2n}\right)-2\sin\left(\frac{\pi}{2n}\right)\sum_{j=0}^{n-2}\left\lfloor\frac{(j+1)^2}{n}\right\rfloor\cos\left(\frac{(j+1)\pi}{n}\right)\\ &=(n-1)\sin\left(\frac{\pi}{2n}\right)-2\sin\left(\frac{\pi}{2n}\right)\underbrace{\sum_{j=1}^{n-1}\left\lfloor\frac{j^2}{n}\right\rfloor\cos\left(\frac{j\pi}{n}\right)}_{=A}\end{align*}
With sum A, using reverse summand property we get
\begin{align*}A&=\sum_{j=1}^{n-1}\left\lfloor\frac{j^2}{n}\right\rfloor\cos\left(\frac{j\pi}{n}\right)\\ &=\sum_{j=1}^{n-1}\left\lfloor\frac{(n-j)^2}{n}\right\rfloor\cos\left(\frac{(n-j)\pi}{n}\right)\\ &=-\sum_{j=1}^{n-1}\left\lfloor n-2j+\frac{j^2}{n}\right\rfloor\cos\left(\frac{j\pi}{n}\right)\\ &=-A+\sum_{j=1}^{n-1}(2j-n)\cos\left(\frac{j\pi}{n}\right)\\ \Rightarrow A&=\frac{1}{2}\sum_{j=1}^{n-1}(2j-n)\cos\left(\frac{j\pi}{n}\right)\end{align*}
We get
$\displaystyle\cos\left(\frac{j\pi}{n}\right)=\dfrac{1}{2\sin\left(\frac{\pi}{2n}\right)}\left[\sin\left(\frac{(2j+1)\pi}{2n}\right)-\sin\left(\frac{(2j-1)\pi}{2n}\right)\right]=\dfrac{1}{2\sin\left(\frac{\pi}{2n}\right)}\Delta\left[\sin\left(\frac{(2j-1)\pi}{2n}\right)\right]$
continue using 1 :)
$\displaystyle A =\left.\dfrac{(2j-n)}{4\sin\left(\frac{\pi}{2n}\right)}\sin\left(\frac{(2j-1)\pi}{2n}\right)\right|_{j=1}^{n}-\dfrac{1}{4\sin\left(\frac{\pi}{2n}\right)}\sum_{j=1}^{n-1}2\sin\left(\frac{(2j+1)\pi}{2n}\right)$
\begin{align*}A&=\frac{n-1}{2}+\dfrac{1}{4\sin^2\left(\frac{\pi}{2n}\right)}\sum_{j=1}^{n-1}\Delta\left[\cos\left(\frac{j\pi}{n}\right)\right]\\ &=\frac{n-1}{2}+\dfrac{1}{4\sin^2\left(\frac{\pi}{2n}\right)}\cdot\left.\cos\left(\frac{j\pi}{n}\right)\right|_{j=1}^n\\ &=\frac{n-1}{2}-\dfrac{1+\cos\left(\frac{\pi}{n}\right)}{4\sin^2\left(\frac{\pi}{2n}\right)}\\&=\frac{n-1}{2}-\dfrac{2\cos^2\left(\frac{\pi}{2n}\right)}{4\sin^2\left(\frac{\pi}{2n}\right)}\end{align*}
Therefore:
\begin{align*}S&=(n-1)\sin\left(\frac{\pi}{2n}\right)-2\sin\left(\frac{\pi}{2n}\right)A\\ &=(n-1)\sin\left(\frac{\pi}{2n}\right)-(n-1)\sin\left(\frac{\pi}{2n}\right)+\dfrac{\cos^2\left(\frac{\pi}{2n}\right)}{\sin\left(\frac{\pi}{2n}\right)}\\ &=\boxed{\displaystyle\cot\left(\frac{\pi}{2n}\right)\cos\left(\frac{\pi}{2n}\right)}\end{align*}
