Find all sequences $(a_n)_{n\geq 1}$ of positive integers such that for all integers $n\geq 3$ we have
\[\dfrac{1}{a_1 a_3} + \dfrac{1}{a_2a_4} + \cdots + \dfrac{1}{a_{n-2}a_n}= 1 – \dfrac{1}{a_1^2+a_2^2+\cdots +a_{n-1}^2}.\]
求所有的正整数列$(a_n)_{n\geq 1}$,满足对一切正整数$n\geq 3$都有\[\dfrac{1}{a_1 a_3} + \dfrac{1}{a_2a_4} + \cdots + \dfrac{1}{a_{n-2}a_n}= 1 – \dfrac{1}{a_1^2+a_2^2+\cdots +a_{n-1}^2}.\]

考虑$n=3$,可得\[    \frac{1}{a_1 a_3}=1-\frac{1}{a_1^2+a_2^2},a_1 a_3=\frac{a_1^2+a_2^2}{a_1^2+a_2^2-1}\]
所以必须有$a_1^2+a_2^2-1=1,a_1=a_2=1,a_3=2$
考虑\begin{gather}    \dfrac{1}{a_1 a_3} + \dfrac{1}{a_2a_4} + \cdots + \dfrac{1}{a_{n-2}a_n}    = 1 – \dfrac{1}{a_1^2+a_2^2+\cdots +a_{n-1}^2}\\    \dfrac{1}{a_1 a_3} + \dfrac{1}{a_2a_4} + \cdots + \dfrac{1}{a_{n-1}a_{n+1}}    = 1 – \dfrac{1}{a_1^2+a_2^2+\cdots +a_{n-1}^2+a_n^2}\\    \frac{1}{a_{n-1}a_{n+1}}=\dfrac{1}{a_1^2+a_2^2+\cdots +a_{n-1}^2}    -\dfrac{1}{a_1^2+a_2^2+\cdots +a_{n-1}^2+a_n^2}\\    (a_1^2+a_2^2+\cdots +a_{n-1}^2)(a_1^2+a_2^2+\cdots +a_{n}^2)=a_{n-1} a_n^2 a_{n+1}\\    \frac{a_1^2+a_2^2+\cdots +a_{n-1}^2}{a_{n-1} a_n}     \frac{a_1^2+a_2^2+\cdots +a_{n}^2}{a_{n} a_{n+1}}=1 \end{gather}
记$b_n=\frac{a_1^2+a_2^2+\cdots +a_{n}^2}{a_{n} a_{n+1}},b_{n-1}b_n=1,n \ge 3$
结合$b_1=b_2=1$,可得对一切$n \in \mN$都有$b_n=1$\begin{gather}a_1^2+a_2^2+\cdots +a_{n}^2=a_{n} a_{n+1}\\a_1^2+a_2^2+\cdots +a_{n+1}^2=a_{n+1} a_{n+2}\\a_{n+1}^2=a_{n+1} a_{n+2}-a_{n} a_{n+1}\\a_{n+2}=a_{n+1}+a_n\end{gather}
所以$\{a_n\}$是斐波那契数列,通项为\[    a_n=\frac{1}{\sqrt{5}}\left((\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n\right)\]

对于一个正$n$边形,各边斜率均存在且分别为$k_j,j=1,2,\dots,n$,如果某边的倾斜角为$\theta$,化简$\ds A_n=\sum_{j=1}^n k_j\sum_{j=1}^n \frac{1}{k_j}$.

昨晚睡前在QQ群看到的一个讨论,原题是$n=5$,推广到$n$.和两年前写的下面这篇”从高考题到巴塞尔问题”关联比较大,如果没有接触过此类问题,可以从下面的文章开始阅读.

从高考题到巴塞尔问题

下面先证明一个引理:\[    F_n(\theta)=\sum_{j=1}^n \tan(\theta+\frac{j\pi}{n})=\begin{cases}        n\tan n\theta & 2 \nmid n\\        -n\cot n\theta & 2 | n    \end{cases}\]
证明:
考虑$x\in \{\theta+\frac{j\pi}{n} |j=1,2,\dots,n\}$的共同之处:$\tan nx=\tan n\theta$
且显然所有$\tan x$是互不相同的
考虑棣莫弗公式\[    \cos nx+\mi \sin nx=(\cos x+ \mi \sin x)^n\]
如果$n=2k+1$是奇数,比较实部虚部可得\begin{gather}    \tan n\theta=\tan nx=\frac{\sin nx}{\cos nx}\\    =\frac{C_{2k+1}^{2k+1}(-1)^k \sin^{2k+1} x+\dots}{C_{2k+1}^{2k}(-1)^k \cos x\sin^{2k} x+\dots}    =\frac{C_{2k+1}^{2k+1}(-1)^k \tan^{2k+1} x+\dots}{C_{2k+1}^{2k}(-1)^k \tan^{2k} x+\dots}\\    C_{2k+1}^{2k+1}(-1)^k \tan^{2k+1} x – \tan (2k+1)\theta \cdot C_{2k+1}^{2k}(-1)^k \tan^{2k} x+\dots=0\end{gather}从而$\tan(\theta+\frac{j\pi}{n})$是如下关于$y$的$2k+1$次方程的$2k+1$个不同根\[    C_{2k+1}^{2k+1}(-1)^k y^{2k+1} – \tan (2k+1)\theta \cdot C_{2k+1}^{2k}(-1)^k y^{2k}+\dots=0\]
韦达定理可得这些根的和\[    F_n(\theta)=-\frac{- \tan (2k+1)\theta \cdot C_{2k+1}^{2k}(-1)^k}{C_{2k+1}^{2k+1}(-1)^k} =(2k+1)\tan (2k+1)\theta=n\tan n\theta\]
如果$n=2k$是偶数,那么按照同样的操作同理可得$\tan(\theta+\frac{j\pi}{n})$是如下关于$y$的$2k$次方程的$2k$个不同根\[    C_{2k}^{2k} (-1)^k y^{2k} – \cot 2k \theta \cdot C_{2k}^{2k-1}(-1)^{k-1} y^{2k-1}+\dots=0\]
韦达定理可得\[    F_n(\theta)=-\frac{- \cot 2k \theta \cdot C_{2k}^{2k-1}(-1)^{k-1}}{C_{2k}^{2k} (-1)^k} =-2k\cot 2k\theta=-n\cot n\theta\]
综上可知\[    F_n(\theta)=\sum_{j=1}^n \tan(\theta+\frac{j\pi}{n})=\begin{cases}        n\tan n\theta & 2 \nmid n\\        -n\cot n\theta & 2 | n    \end{cases}\]

接着证明原题:
不妨设第$j$边倾斜角为$x_j=\theta+\frac{2j\pi}{n}$,(不管倾斜角$[0,\pi)$的取值范围,求正切以后无影响)\[    f_n(\theta)=\sum_{j=1}^n \tan(\theta+\frac{2j\pi}{n})\]
则\[    \sum_{j=1}^n \cot(\theta+\frac{2j\pi}{n})=\sum_{j=1}^n \tan(\frac{\pi}{2}-\theta-\frac{2j\pi}{n})=f_n(\frac{\pi}{2}-\theta)\]
待求式即为$A_n=f_n(\theta)f_n(\frac{\pi}{2}-\theta)$
若$n=2k+1$是奇数,则易得任意两个边倾斜角的差都不是$\pi$的整数倍,所以求正切后互不相同,是$\theta+\frac{j\pi}{n}$的重新排列
亦即$f_n(\theta)=F_n(\theta)$
\[    f_n(\frac{\pi}{2}-\theta)=n \tan n(\frac{\pi}{2}-\theta)=n \tan (k\pi+\frac{\pi}{2}-n\theta)=n \cot n\theta\]
于是待求为\[    A_n=f_n(\theta)f_n(\frac{\pi}{2}-\theta)=\left( n\tan n\theta \right) \left( n \cot n\theta \right)=n^2\]
若$n=2k$是偶数,则$x_1,x_2,\dots,x_k$的正切刚好取遍$\tan(\theta+\frac{j\pi}{k}),j=1,2,\dots,k$,而$\tan x_{k+j}=\tan x_j$
所以$f_{2k}(\theta)=2F_k(\theta),f_{2k}(\frac{\pi}{2}-\theta)=2F_k(\frac{\pi}{2}-\theta)$,待求式为\[    A_n=f_{2k}(\theta) f_{2k}(\frac{\pi}{2}-\theta)=4F_k(\theta)F_{k}(\frac{\pi}{2}-\theta)\]
如果$k$是奇数,如上可得\[    A_n=4k^2=n^2    \]
如果$k=2m$是偶数,那么$n=4m$\[    A_n=4F_{2m}(\theta)F_{2m}(\frac{\pi}{2}-\theta)=4 \cdot (-2m\cot 2m\theta) \cdot (-2m \cot 2m(\frac{\pi}{2}-\theta))=-n^2\cot^2\frac{n\theta}{2}\]

综上可得\[    A_n=\begin{cases}        -n^2\cot^2\frac{n\theta}{2} & 4|n\\        n^2 & else    \end{cases}\]