一类三角函数分式的封闭和

青青子衿

\[ \Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\cos\left(\frac{2k+1}{2n}\pi\right)}=\frac{n}{\sqrt{p^2-q^2}}\tanh\left(n\operatorname{artanh}\frac{\sqrt{p^2-q^2}}{p}\right)} \]
\[ \Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\sin\left(\frac{2k+1}{2n}\pi\right)}=\,???} \]

青青子衿

回复 1# 青青子衿
\begin{align*}  
\Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\cos\left(\frac{2k+1}{n}\pi\right)}}
&
\Large\color{black}{=\dfrac{n}{\sqrt{p^{\overset{\,}2}-q^2}}\cdot\dfrac{\left(-q\right)^{n}-\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^{n}\,}{\left(-q\right)^{n}+\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^{n}\,}}\\
\Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\cos\left(\frac{2k+1}{2n}\pi\right)}}
&
\Large\color{black}{=\dfrac{n}{\sqrt{p^{\overset{\,}2}-q^2}}\cdot\dfrac{q^{2n}-\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^{2n}\,}{q^{2n}+\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^{2n}\,}}\\
\Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\cos\left(\frac{2k}{n}\pi\right)}}
&
\Large\color{black}{=\dfrac{n}{\sqrt{p^{\overset{\,}2}-q^2}}\cdot\dfrac{\left(-q\right)^{n}+\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^{n}\,}{\left(-q\right)^{n}-\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^{n}\,}}\\
\Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\cos\left(\frac{k}{n}\pi\right)}}
&
\Large\color{black}{=\dfrac{n}{\sqrt{p^{\overset{\,}2}-q^2}}\cdot\dfrac{q^{2n}+\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^{2n}\,}{q^{2n}-\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^{2n}\,}-\dfrac{q}{p^2-q^2}}\\
\end{align*}
...

1. DESMOS code
2. \sum_{k=0}^{n-1}\frac{1}{p+q\cos\left(\frac{2k+1}{n}\pi\right)}
3. \frac{n}{\sqrt{p^2-q^2}}\cdot\frac{\left(-q\right)^n-\left(p-\sqrt{p^2-q^2}\right)^n}{\left(-q\right)^n+\left(p-\sqrt{p^2-q^2}\right)^n}
4. \sum_{k=0}^{n-1}\frac{1}{p+q\cos\left(\frac{2k+1}{2n}\pi\right)}
5. \frac{n}{\sqrt{p^2-q^2}}\cdot\frac{q^{2n}-\left(p-\sqrt{p^2-q^2}\right)^{2n}}{q^{2n}+\left(p-\sqrt{p^2-q^2}\right)^{2n}}
6. \sum_{k=0}^{n-1}\frac{1}{p+q\cos\left(\frac{2k}{n}\pi\right)}
7. \frac{n}{\sqrt{p^2-q^2}}\cdot\frac{\left(-q\right)^n+\left(p-\sqrt{p^2-q^2}\right)^n}{\left(-q\right)^n-\left(p-\sqrt{p^2-q^2}\right)^n}
8. \sum_{k=0}^{n-1}\frac{1}{p+q\cos\left(\frac{k}{n}\pi\right)}
9. \frac{n}{\sqrt{p^2-q^2}}\cdot\frac{q^{2n}+\left(p-\sqrt{p^2-q^2}\right)^{2n}}{q^{2n}-\left(p-\sqrt{p^2-q^2}\right)^{2n}}-\frac{q}{p^2-q^2}

Copy the Code

...
\begin{align*}
\Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\cos\left(\frac{2k+1}{n}\pi\right)}}  
&  
\Large\color{black}{=\begin{cases}\dfrac{n}{\sqrt{p^{\overset{\,}2}-q^2}}\coth\left(\dfrac{n}{2}\operatorname{artanh}\dfrac{\sqrt{p^{\overset{\,}2}-q^2}}{p}\right) & n= 1\pmod2 \\
\dfrac{n}{\sqrt{p^{\overset{\,}2}-q^2}}\tanh\left(\dfrac{n}{2}\operatorname{artanh}\dfrac{\sqrt{p^{\overset{\,}2}-q^2}}{p}\right)& n= 0\pmod2\end{cases}}\\  
\,\\
\Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\cos\left(\frac{2k+1}{2n}\pi\right)}}  
&  
\Large\color{black}{=\dfrac{n}{\sqrt{p^{\overset{\,}2}-q^2}}\tanh\left(n\operatorname{artanh}\dfrac{\sqrt{p^{\overset{\,}2}-q^2}}{p}\right)}\\  
\,\\
\Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\cos\left(\frac{2k}{n}\pi\right)}}  
&  
\Large\color{black}{=\begin{cases}\dfrac{n}{\sqrt{p^{\overset{\,}2}-q^2}}\tanh\left(\dfrac{n}{2}\operatorname{artanh}\dfrac{\sqrt{p^{\overset{\,}2}-q^2}}{p}\right) & n= 1\pmod2 \\
\dfrac{n}{\sqrt{p^{\overset{\,}2}-q^2}}\coth\left(\dfrac{n}{2}\operatorname{artanh}\dfrac{\sqrt{p^{\overset{\,}2}-q^2}}{p}\right)& n= 0\pmod2\end{cases}}\\  
\,\\
\Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\cos\left(\frac{k}{n}\pi\right)}}  
&  
\Large\color{black}{=\dfrac{n}{\sqrt{p^{\overset{\,}2}-q^2}}\coth\left(n\operatorname{artanh}\dfrac{\sqrt{p^{\overset{\,}2}-q^2}}{p}\right)-\dfrac{q}{p^2-q^2}}
\end{align*}
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青青子衿

回复 1# 青青子衿
\begin{align*}   
\Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\sin\left(\frac{2k+1}{n}\pi\right)}}  
&  
\Large\color{black}{=\dfrac{n}{\sqrt{p^{\overset{\,}2}-q^2}}\cdot\dfrac{q^{2n}-\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^{2n}\,}{q^{2n}+2\left(-q\right)^n\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^n\cos\left(\frac{n}{2}\pi\right)+\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^{2n}\,}}\\  
\Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\sin\left(\frac{2k+1}{2n}\pi\right)}}  
&  
\Large\color{black}{=\,?????}\\  
\Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\sin\left(\frac{2k}{n}\pi\right)}}  
&  
\Large\color{black}{=\dfrac{n}{\sqrt{p^{\overset{\,}2}-q^2}}\cdot\dfrac{q^{2n}-\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^{2n}\,}{q^{2n}-2\left(-q\right)^n\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^n\cos\left(\frac{n}{2}\pi\right)+\left(p-\sqrt{p^{\overset{\,}2}-q^2}\right)^{2n}\,}}\\  
\Large\color{black}{\sum_{k=0}^{n-1}\frac{1}{p+q\sin\left(\frac{k}{n}\pi\right)}}  
&  
\Large\color{black}{=\,?????}\\  
\end{align*}