Adventure with infinite series, a curiosity

It is moderately nice, I would say. We have $\sum \binom{2k}k x^k=(1-4x)^{-1/2}$ for $|x|<1/4$. If we need only terms with $k$ divisible by 4 and $x=2^{-5/2}$, $4x=2^{-1/2}$, we get $$\sum \binom{8k}{4k}2^{-10k}=\frac14\sum_{w^4=1}(1-w/\sqrt{2})^{-1/2}$$ and so on.

More generally, it seems that $$ \sum_{k\ge0}\binom{2^{j+1}k}{2^jk}2^{-a_{j+2} k}=2^{-j}\sum_{w^{2^j}=1}(1-w/\sqrt2)^{-1/2} $$ where $a_1=2,a_2=3,a_{j+1}=a_j+a_{j-1}+\dots+a_1$ (cf. A257113).

Let $$f(x):=\sum_{k\ge 0}\binom{2k}{k}x^k=\frac{1}{\sqrt{1-4x}}$$ with $|x|<1/4$. We have for $N,j\in\mathbb{Z}_+$, \begin{align} \frac{1}{N}\sum_{r=1}^Ne(-{jr}/{N})f\left(xe({r}/{N})\right)&=\frac{1}{N}\sum_{r=1}^Ne(-{jr}/{N})\sum_{k\ge 0}\binom{2k}{k}x^ke({kr}/{N})\\ &=\sum_{k\ge 0}\binom{2k}{k}x^k\frac{1}{N}\sum_{r=1}^Ne({(k-j)r}/{N}), \end{align} where $e(x):=e^{2\pi{\rm i}x}$. Hence clearly, $$\sum_{\substack{k\ge 0\\ k\equiv j\pmod N}}\binom{2k}{k}x^k=\frac{1}{N}\sum_{r=1}^N\frac{e(-{jr}/{N})}{\sqrt{1-4xe(r/N)}},$$ holds for all $N,j\in\mathbb{Z}_+$, which is a more general result.