Interpolating a sum of binomial coefficients using a sin function While studying a problem about orthogonal polynomials I encountered the following
 expressions
\begin{equation}
f(n)=\sum_{k=0}^{n}(-1)^k\binom{n+k}{2k} \frac{1}{k+1}\binom{2k}{k}
\end{equation}
and
\begin{equation}
g(n)=\sum_{k=0}^{n-1}(-1)^k\binom{n+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}
\end{equation}
I can prove that $f(n)=0$ for all integers $n\geq 1$ and $g(n)=0$ for all integers $n>1$ using properties of orthogonal polynomials. However,  I would like  to find an elementary proof, which might also  be more illuminating.
While doing some experiments with calculations using Mathematica, I defined the functions $f(n)$ and $g(n)$ with the above formulas but forgot to specify that $n$ is an integer value.
When I typed $f(n)$ and $g(n)$ for a generic variable $n$, I guess that Mathematica "assumed" that the variables involved were real and provided the following simple formulas:
\begin{equation}
f(x)=\frac{\sin \pi x}{x(x+1)\pi}
\end{equation}
and
\begin{equation}
g(x)=-\frac{2\sin \pi x}{(x+1)(x-1)\pi}
\end{equation}
which makes it evident that these functions are zero for all positive (and negative) integers with the possible exceptions of $0,1,-1$.
Why are these equalities true?
I realize it might have to do, perhaps, with properties of the Euler Beta and Gamma functions, but I know too little about these functions to figure out a proof along these lines. Can anybody help? Thank you!
 A: The two identities are both special cases of Vandermonde's theorem (also called the Chu-Vandermonde theorem), which is the most well-known binomial coefficient identity after the binomial theorem.
The first sum may be written
$$f(n) = \frac 1n \sum_{k=0}^n (-1)^k\binom nk \binom {n+k}{k+1}.$$
Applying the identity  $\binom{-c}m = (-1)^{m}\binom{c+m-1}{m}$, we get
$$f(n)= -\frac 1n \sum_{k=0}^n \binom nk \binom {-n}{k+1},$$
and reversing the order of summation gives
\begin{equation*}
f(n) =-\frac 1n \sum_{k=0}^n \binom nk \binom {-n}{n-k+1}.
\end{equation*}
Vandermonde's theorem is often written
\begin{equation*}
\sum_{k=0}^m \binom ak \binom b{m-k} = \binom{a+b}m.
\end{equation*}
There are many simple proofs of Vandermonde's theorem. For example, we can prove this formula by equating coefficients of $x^m$ in $(1+x)^a (1+x)^b = (1+x)^{a+b}$. 
Setting $a=n$, $b=-n$, and $m=n+1$ in this formula (and observing that the $k=n+1$ term vanishes) and using the last formula for $f(n)$ shows that for $n>0$, 
$f(n) =- \frac 1n\binom 0{n+1}=0.$
Similarly, $g(n)$ can be evaluated by Vandermonde's theorem.
I believe that Pietro's indefinite summation approach is correct, but I don't think that this is what Mathematica is doing. There is a nonterminating generalization of Vandermonde's theorem, called Gauss's theorem (see, e.g., http://mathworld.wolfram.com/GausssHypergeometricTheorem.html) that evaluates the sum $\sum_{k=0}^\infty (-1)^k \binom ak \binom {b+k}{m+k}$ in terms of gamma functions when it converges, where $a$ and $b$ are arbitrary and $m$ is an integer. (Actually Gauss's theorem is a little more general than this.) In the particular case $a=n$, $b=n$, $m=1$, the reflection formula for the gamma function can be applied to give Stefano's formula for $f(x)$ in terms of $\sin \pi x$, and similarly for $g(x)$. Mathematica is probably applying Gauss's theorem, since it knows how to convert binomial coefficient sums to hypergeometric series and it knows how to evaluate them in cases like this one.
A: Let's consider more generally the sums (from $0$ to $n$, resp. from $0$ to $n-1$) with a real or complex number $x$ in place of the $n$ inside the sums:  
\begin{equation}
\sum_{k=0}^{n}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k}
\end{equation}
and
\begin{equation}
\sum_{k=0}^{n-1}(-1)^k\binom{x+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}\, .
\end{equation}
These sums can be evaluated by induction as:
\begin{equation}
(-1)^n\frac{n+1}{x(x+1)} \binom{x+n+1}{2n+2}\binom{2n+2}{n+1}
\end{equation}
and
\begin{equation}
(-1)^n\frac{n+1}{x^2-1} \binom{x+n+1}{2n+3}\binom{2n+4}{n+2}\, .
\end{equation}
We can write these expression respectively as
\begin{equation}
\frac{1}{x+1}\left(1+\frac{x}{n+1} \right)\prod_{k=1}^n\left(1-\frac{x^2}{k^2}\right)
\end{equation}
and
\begin{equation}
-\frac{2x}{x^2-1}\frac{n+1}{n+2}\prod_{k=1}^{n+1}\left(1-\frac{x^2}{k^2}\right)\, .
\end{equation}
The limit of these expression can be evaluated by means of the Euler's infinite product for $\sin(\pi x)$, obtaining respectively the values of the series :
\begin{equation}
\sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k}=\frac{\sin(\pi x)}{\pi x(x+1)}
\end{equation}
and
\begin{equation}
\sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}=-\frac{2\sin(\pi x)}{\pi(x^2-1)}\, .
\end{equation}
As observed in the comments, for a positive integer $x=n$, these series coincide respectively with the initially considered finite sums $f$ and $g$, of which they may be considered therefore as a natural extension to complex values of $x$. 
