For $n \geq 0$ evenly distribute $2n$ points on the circumference of a circle, and label these point cyclically with the numbers $1, 2 . . . , 2n$ Let $h_n$ be the number of ways in which these $2n$ points can be paired off as $n$ chords where no two chords intersect I want […]

Consider $\varphi=\frac{1+\sqrt{5}}{2}$, the golden ratio. Bellow are series $(3)$ and $(6)$ that represent $\varphi$ $$ \begin{align*} \varphi &=\frac{1}{1}+\sum_{k=0}^{\infty}\cdots&(1)\\ \varphi &=\frac{2}{1}+\sum_{k=0}^{\infty}\cdots&(2)\\ \varphi &=\frac{3}{2}+\sum_{k=0}^{\infty}(-1)^{k}\frac{(2k)!}{(k+1)!k!2^{4k+3}}&(3)\\ \varphi &=\frac{5}{3}+\sum_{k=0}^{\infty}\cdots&(4)\\ \varphi &=\frac{8}{5}+\sum_{k=0}^{\infty}\cdots&(5)\\ \varphi &=\frac{13}{8}+\sum_{k=0}^{\infty}(-1)^{k+1}\frac{(2(k+1))!}{((k+1)+1)!(k+1)!2^{4(k+1)+3}}&(6)\\ \vdots&\\ \end{align*} $$ When looking at the leading terms of $(3)$ and $(6)$ $\;\frac{3}{2}$ and $\frac{13}{8}$ respectively, one is tempted to conjecture that there are similar formulas […]

I’m stuck with the proof of this result: $$2^n = \sum_{t=-\frac{n-1}{2}}^{\frac{n-1}{2}} \binom{n+1}{\frac{n+1}{2} + t} \sum_{k=\vert t \vert}^{\frac{n-1}{2}} \binom{\frac{n-1}{2}+k}{k} \binom{2k}{k+t} \frac{(-1)^t}{2^{2k}}$$ where $n$ is an odd integer. How can we deal with this kind of summation? Thanks in advance.

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