TITLE: Combinatorial identity with sum of binomial coefficients
QUESTION [17 upvotes]: How to attack this kinds of problem? I am hoping that there will some kind of shortcuts to calculate this.   
$$\sum_{k=0}^{38\,204\,629\,939\,869} \frac{\binom{38\,204\,629\,939\,869}{k}}{\binom{76\,409\,259\,879\,737}{k}}\,.$$
EDIT:
As I see, the numerator is $n \choose k$ and the denominator is ${2n-1} \choose k$, where $n =38\,204\,629\,939\,869$. i.e $$\sum_{k=0}^n {\frac {n \choose k} {{2n-1} \choose k}} = 2.$$

REPLY [2 votes]: According to the Gosper's algorithm (Maxima command
AntiDifference(binomial(n,k)/binomial(2*n-1,k),k),
also implemented  in Mathematica and Maple):
$$
{\frac {n \choose k} {{2n-1} \choose k}}
=
{{\left((k+1)-2n\right){{n}\choose{k+1}}}\over{n{{2n-1}\choose{k+1
 }}}}
-{{\left(k-2n\right){{n}\choose{k}}}\over{n{{2n-1}\choose{k}}}}
$$
and the sum telescopes :
$$
\sum_{k=0}^n{\frac{n \choose k}{{2n-1} \choose k}}
=
\sum_{k=0}^n{{\left((k+1)-2n\right){{n}\choose{k+1}}}\over{n{{2n-1}\choose{k+1}}}}
-{{\left(k-2n\right){{n}\choose{k+1}}}\over{n{{2n-1}\choose{k}}}}=
{{\left(1-n\right){{n}\choose{n+1}}}\over{n{{2n-1}\choose{n}}}}-
{{\left(-2n\right){{n}\choose{0}}}\over{n{{2n-1}\choose{0}}}}=0-(-2)
$$