<s>[INST] Adventure with infinite series, a curiosity 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.
 [/INST] Mathematica says, for the first:
$$\,
   _4F_3\left(\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8};\frac{1}{4},\frac{1}{2},\frac{
   3}{4};\frac{1}{4}\right)$$
and
$$\,
   _8F_7\left(\frac{1}{16},\frac{3}{16},\frac{5}{16},\frac{7}{16},\frac{9}{16},\frac{11}{16}
   ,\frac{13}{16},\frac{15}{16};\frac{1}{8},\frac{1}{4},\frac{3}{8},\frac{1}{2},\frac{5}{8},
   \frac{3}{4},\frac{7}{8};\frac{1}{16}\right)
$$ for the second.
</s>