Combinatorial identity with sum of binomial coefficients 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 $$\sum_{k=0}^n {\frac {n \choose k} {{2n-1} \choose k}} = 2.$$

The identity
$$\sum_{k=0}^n\frac{\binom nk}{\binom{2n-1}k}=2$$
holds for every positive integer $n$. The case $n=1$ is trivial. Here is a probabilistic proof for $n\ge2$.
Consider the following random experiment. An urn initially contains $n$ black balls and $n-1$ white balls. Balls are drawn one by one, without replacement, until a white ball is drawn. The random variable $X$ is the number of draws; its range of values is $\{1,2,\dots,n,n+1\}$. We will compute the expected value $E(X)$ in two different ways.
I. Clearly we have
$$X=\sum_{k=0}^nX_k$$
where
$$X_k=\begin{cases}
1\text{ if }X\gt k,\\
0\text{ if }X\le k;
\end{cases}$$
in other words, $X_k=1$ if there is no white ball in the first $k$ draws, meaning that a $(k+1)^{\text{st}}$ draw is needed. Thus we have
$$E(X)=E(\sum_{k=0}^n X_k)=\sum_{k=0}^n E(X_k)=\sum_{k=0}^n P(X_k=1)=\sum_{k=0}^n\frac{\binom nk}{\binom{2n-1}k}.$$
II. Call the black balls $B_1,B_2,\dots,B_n$. Let $Y_i$ be the indicator variable which takes the value $1$ if the ball $B_i$ is drawn before any white ball is drawn, $0$ otherwise. Clearly the variable $X$ is equal to $1$ plus the number of black balls drawn, that is,
$$X=1+\sum_{i=1}^nY_i$$
and so
$$E(X)=1+\sum_{i=1}^nE(Y_i)=1+\sum_{i=1}^nP(Y_i=1)=1+\sum_{i=1}^n\frac1n=2.$$
More generally, for any integers $m,n\ge0$, the same argument (with $n$ black and $m$ white balls) shows that
$$\sum_{k=0}^n\frac{\binom nk}{\binom{m+n}k}=1+\frac n{m+1}.$$
