Infinite binomial series over a column of Pascal's triangle: $F_k(x)= \sum\limits_{n=0}^\infty \binom{2n+k}n x^{2n+k}$ Is there any closed form formula (or an equivalent) for this binomial infinite series :
$$F_k(x)=  \sum_{n=0}^{\infty} \binom{2n+k}{ n } x^{2n+k} $$ in which $|x|<1$ and $k$ is a given integer?
This is a sum over a (odd or even) infinitely long column of Pascal's triangle :

For example, above : $$F_3(x)= x^3+ 5 x^5 + 21 x^7 + 84 x^9 +... $$
$F_k(x)$ appears in the Fourier Series development of $\frac{1}{1+a\,sin(x)}$ : the k-th harmonic would be of amplitude $F_k(a)$

We have
$$
S(k,x) = \sum\limits_{n = 0}^\infty  {\left( \matrix{  2n + k \cr   n \cr}  \right)x^{2n + k} }
  = x^k \sum\limits_{n = 0}^\infty  {\left( \matrix{  2n + k \cr   n \cr}  \right)\left( {x^2 } \right)^n } 
$$
Indicating the series coefficients as
$$
t_n  = \left( \matrix{  2n + k \cr   n \cr}  \right)
$$
then we have
$$
\eqalign{
  & t_0  = 1  \cr   & {{t_{n + 1} } \over {t_n }} =
 {{\left( \matrix{  2n + 2 + k \cr   n + 1 \cr}  \right)} \over {\left( \matrix{  2n + k \cr   n \cr}  \right)}} =
 {{n!\left( {n + k} \right)!\left( {2n + 2 + k} \right)!} \over {\left( {n + 1} \right)!\left( {n + 1 + k} \right)!\left( {2n + k} \right)!}}
 = {{\left( {2n + 2 + k} \right)\left( {2n + 1 + k} \right)} \over {\left( {n + 1} \right)\left( {n + 1 + k} \right)}} =   \cr 
  &  = 4{{\left( {n + 1 + k/2} \right)\left( {n + 1/2 + k/2} \right)} \over {\left( {n + 1} \right)\left( {n + 1 + k} \right)}} \cr} 
$$
which means that the series might be expressed through a
Hypergeometric function
$$
\eqalign{
  & S(k,x) = x^k \sum\limits_{0\, \le \;k} {{{\left( {1/2 + k/2} \right)^{\,\overline {\,n\,} }
 \left( {1 + k/2} \right)^{\,\overline {\,n\,} } } \over {\left( {1 + k} \right)^{\,\overline {\,n\,} } }}
{{\left( {4x^2 } \right)^{\,n} } \over {n!}}}  =   \cr 
  &  = x^k {}_2F_{\,1} \left( {\left. {\matrix{ {1/2 + k/2,\;1 + k/2}  \cr {1 + k}  \cr } \;}
 \right|\;4x^2 } \right) \cr} 
$$
which converges for $|x|<1/2$
To explain the convergence range, it is known that
$$
S(0,x) = \sum\limits_{n = 0}^\infty  {\left( \matrix{  2n \cr   n \cr}  \right)x^{2n} }  =
 {1 \over {\sqrt {1 - 4x^2 } }}
$$
The duplication formula for Gamma in fact gives
$$
\eqalign{
  & \left( \matrix{
  2n \cr 
  n \cr}  \right) = {{\Gamma \left( {2n + 1} \right)} \over {\Gamma \left( {n + 1} \right)^2 }} =
 {{4^{\,n} } \over {\sqrt \pi  }}{{\Gamma \left( {n + 1/2} \right)} \over {\Gamma \left( {n + 1} \right)}} =   \cr 
  &  = 4^{\,n} {{\Gamma \left( {n + 1/2} \right)} \over {\Gamma \left( {1/2} \right)\Gamma \left( {n + 1} \right)}} =
 4^{\,n} \left( \matrix{  n - 1/2 \cr   n \cr}  \right) =   \cr 
  &  = \left( { - 4} \right)^{\,n} \left( \matrix{ - 1/2 \cr  n \cr}  \right) \cr} 
$$
