It is easily verifiable that
$$\sum\_{k\geq0}\binom{2k}k\frac1{2^{3k}}=\sqrt{2}.$$
It is not that difficult to get
$$\sum\_{k\geq0}\binom{4k}{2k}\frac1{2^{5k}}=\frac{\sqrt{2-\sqrt2}+\sqrt{2+\sqrt2}}2.$$



> 
> **Question.** Is there something similarly "nice" in computing
>  $$\sum\_{k\geq0}\binom{8k}{4k}\frac1{2^{10k}}=?$$
>  Perhaps the same question about
>  $$\sum\_{k\geq0}\binom{16k}{8k}\frac1{2^{20k}}=?$$
> 
> 
> 


**NOTE.** The powers of $2$ are selected with a hope (suspicion) for some pattern.



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.


