Proof: $\sum\limits_{r=0}^{\infty}{n\choose t+rs}=\frac1s\sum\limits_{j=0}^{s-1} \left(2\cos\frac{\pi j}2\right)^n\cos\frac{\pi j(n-2t)}s$

Question

I found somewhere on the internet (I forget where) the strange identity

For $s,t\in\Bbb Z,\quad 0\leq t<s:$ $$\sum_{r\geq0}{n\choose t+rs}=\frac1s\sum_{j=0}^{s-1}\left(2\cos\frac{\pi j}2\right)^n\cos\frac{\pi j(n-2t)}s$$ Although I do not know the restrictions on $n$, I suspect $n\in\Bbb N$.

I am attempting to prove this curious identity. I start by defining $$F(n;t,s)=\sum_{r\geq0}{n\choose t+rs}$$ I was thinking we could use the integral given by the Wolfram functions Site: $${n\choose k}=\frac1{2\pi}\int_{-\pi}^\pi e^{-ikx}(1+e^{ix})^{n}dx$$ But I am unsure if I would be able validly to interchange the sum and integral signs. Other than that I thought I might be able to use my usual trick of representing the binomial as a beta integral, but I cannot seem to do that in this case. Any help would be appreciated.

Answer 1

The $s$-th roots of unity $\omega_j=\exp\left(\frac{2\pi ij}{s}\right), 0\leq j<s$ have the nice property to filter elements. For $s>0$ we obtain \begin{align*} \frac{1}{s}\sum_{j=0}^{s-1}\exp\left(\frac{2\pi ij n}{s}\right)= \begin{cases} 1&\qquad s\mid n\\ 0& \qquad otherwise \end{cases} \end{align*}

We can use it for series multisection of $f(z)=\sum_{r=0}^\infty a_rz^r$ to obtain \begin{align*} \sum_{r=0}^\infty a_{t+rs}z^{t+rs}=\frac{1}{s}\sum_{j=0}^{s-1}\omega_j^{-t}f(\omega_jz)\qquad s,t\in\mathbb{Z}, s>0, t\geq 0\tag{1} \end{align*}

We apply (1) to $f(z)=(1+z)^n=\sum_{r=0}^n \binom{n}{r}z^r$ and obtain setting $z=1$ \begin{align*} \color{blue}{\sum_{r=0}^n}\color{blue}{\binom{n}{t+rs}}& =\frac{1}{s}\sum_{j=0}^{s-1}\omega_j^{-t}(1+\omega_j)^n\\ &=\frac{1}{s}\sum_{j=0}^{s-1}\exp\left(-\frac{2i\pi jt}{s}\right)\left(1+\exp\left(\frac{2i\pi j}{s}\right)\right)^n\\ &=\frac{1}{s}\sum_{j=0}^{s-1}\exp\left(\frac{i\pi j (n-2t)}{s}\right) \left(\exp\left(-\frac{i\pi j}{s}\right)+\exp\left(\frac{i\pi j}{s}\right)\right)^n\tag{2}\\ &=\frac{1}{s}\sum_{j=0}^{s-1}\left(\cos\left(\frac{\pi j (n-2t)}{s}\right)+i\sin\left(\frac{\pi j(n-2t)}{s}\right)\right)\\ &\qquad \cdot\left(2\cos\left(\frac{\pi j}{s}\right)\right)^n\tag{3}\\ &\,\,\color{blue}{=\frac{1}{s}\sum_{j=0}^{s-1}\left(2\cos\left(\frac{\pi j}{s}\right)\right)^n\cos\left(\frac{\pi j(n-2t)}{s}\right)}\tag{4} \end{align*} and the claim follows.

Comment:

• In (2) we factor out $\exp\left(\frac{\pi nj}{s}\right)$.

• In (3) we use the identities $e^{iz}=\cos(z)+i\sin(z)$ and $\cos(z)=\frac{1}{2}\left(e^{iz}+e^{-iz}\right)$.

• In (4) we skip the imaginary part (which is zero), since we know the result is real-valued.