Some keywords you'll want to look into: binomial type, Appell sequence, Sheffer sequence, umbral calculus. The references in the corresponding Wikipedia articles are good too.
Edit: In some sense, all of these identities can be deduced from the last one. Setting
$$f(t) = e^{xt}, g(t) = e^{yt}$$
produces the binomial theorem, and setting
$$f(t) = (1 + t)^x = \exp (x \log (1 + t)), g(t) = (1 + t)^y = \exp ( y \log (1 + t))$$
produces the second identity. From this perspective one can think of the study of generalized binomial theorems as being all about generating functions of the form $\exp (x h(t))$ where $h(0) = 0$; setting
$$f(t) = \exp (x h(t)), g(t) = \exp (y h(t))$$
produces a fairly general binomial theorem, especially if one writes $h(t) = \sum_{n \ge 1} h_n t^n$ as a formal power series in formal variables.
Generally similar sums can be evaluated using the Beta function:
$$
B(x+1,y+1)=\int_0^1 t^{x}(1-t)^{y}dt=\frac{\Gamma(x+1)\Gamma(y+1)}{\Gamma(x+y+2)}=
\frac{x!y!}{(x+y+1)!}=\frac{1}{x+y+1}\binom{x+y}x^{-1}.
$$
Applying this in your case ($x=n-k,y=n+k$) one has:
$$
\binom{2n}{n-k}^{-1}=(2n+1)\int_0^1 t^{n-k}(1-t)^{n+k}dt
$$
or
$$\begin{align}
\sum_{n=k}^\infty\binom{2n}{n-k}^{-1}
&=\sum_{n=k}^\infty(2n+1)\int_0^1 t^{n-k}(1-t)^{n+k}dt\\
&=\int_0^1 dt \sum_{n=k}^\infty(2n+1)t^{n-k}(1-t)^{n+k}\\
&=\int_0^1 \frac{(1-t)^{2k}[2+(2k-1)(1-t+t^2)]}{(1-t+t^2)^2}dt\tag1
\end{align}$$
It can be shown that the integral $(1)$ is a sum of a rational number and a multiple of $\dfrac\pi{9\sqrt3}$.
Indeed:
$$
\frac{(1-t)^{2k}[2+(2k-1)(1-t+t^2)]}{(1-t+t^2)^2}=Q_k(t)+\frac{A_k^0+A_k^1t+A_k^2t^2+A_k^3t^3}{(1-t+t^2)^2},\tag2
$$
where both coefficients of the polynomial $Q(t)$ and $A^0,A^1,A^2,A^3$ are integer numbers. The integral of $Q(t)$ is obviously a rational number and
$I_r= \int_0^1\frac{t^r\,dt}{(1-t+t^2)^2}$ can be evaluated as:
$$
I_0=\frac23+\frac49\dfrac\pi{\sqrt3};\quad
I_1=\frac13+\frac29\dfrac\pi{\sqrt3};\quad
I_2=-\frac13+\frac49\dfrac\pi{\sqrt3};\quad
I_3=-\frac23+\frac59\dfrac\pi{\sqrt3}.\quad
$$
Thus the irrational term can be written as:
$$
C_k\frac\pi{9\sqrt3}\quad\text{with}\quad C_k=4A_k^0+2A_k^1+4A_k^2+5A_k^3.
$$
Moreover the term can be evaluated explicitly using the following table:
$$
\begin{array}{c|c|c|c|c|c}
k\mod 3& A_k^0&A_k^1&A_k^2&A_k^3&C_k\\
\hline
0&+1-2x&+1+6x&-1-6x&+0+4x&2\\
1&+2+4x&-5-6x&+3+6x&-1-2x&\hphantom{-1}5+18x\\
2&-3-2x&2&0&-1-2x&-13-18x\\
\hline
\end{array},\tag3
$$
with $x=\left\lfloor\dfrac k3\right\rfloor$, so that $C_k=2,5,-13,2,23,-31,2,41,-49,2,\dots$ for $k=0,1,2,3,4,5,6,7,8,9,\dots$.
The expression $(3)$ can be proved in the following way:
Let
$$
P_k(t)=(1-t)^{2k}[2+(2k-1)(1-t+t^2)];\quad R_k(t)=A_k^0+A_k^1t+A_k^2t^2+A_k^3t^3.\\
$$
Then we have from (2):
$$R_k(t_\pm)=P_k(t_\pm);\quad R'_k(t_\pm)=P'_k(t_\pm),\tag4$$
where
$$
t_\pm=e^{\pm\frac{i\pi}3}
$$
are the roots of the polynomial $t^2-t+1$.
Explicitly (4) amounts to the system of four linear equations:
$$\begin{align}
A_k^0+A_k^1t_\pm+A_k^2t_\pm^2+A_k^3t_\pm^3&=2t_\pm^{-2k}\\
A_k^1+2A_k^2t_\pm+3A_k^3t_\pm^2&=(1-2k-2t_\pm)t_\pm^{-2k}\\
\end{align},
$$
which solutions are given by (3).
Best Answer
There is a technique known as Gosper's algorithm and another technique which is known as Zeilberger's algorithm. The two algorithms tackle these kinds of problems, if they succeed they will give a closed form formulas for the finite sum.