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A deleted question (prior answers awaiting merge are viewable here) asked to prove the following:

Let $n$ be a positive integer. Then,
beginalign
sum_k=1^n+1 dfracn binomnk-1binom2 nk=frac2n+1n+1 .
endalign

Note that we can rewrite $dfracn binomnk-1binom2 nk$ as $dfracn!^2 k binom2n-kn-1(2n)!$ (by the standard $dbinomab = dfraca!b!left(a-bright)!$ formula). Thus, the question is equivalent to proving that
beginalign
sum_k=1^n+1 k dbinom2n-kn-1 = dbinom2n+1n+1 .
endalign

The equality $ sum_k=1^n+1 k binom2n-kn-1 = binom2n+1n+1 $ is just a special case of
$$
sum_k=i^m-j binomkibinomm-kj=binomm+1i+j+1.
$$
where $igets1,jgets(n-1),mgets 2n$. For a combinatorial proof:

How many subsets of $1,2,dots,m+1$ have size $i+j+1$?

How many such subsets contain $k+1$, and have exactly $i$ elements smaller than $k+1$?

See $sum_k=-m^n binomm+kr binomn-ks =binomm+n+1r+s+1$ using Counting argument.

Proof Using Convolutions and Generating Functions

First, we show the more general result:

$$
sum_k=0^mbinomklbinomm-kq = binomm+1l+q+1, quad textfor integers m, l, q ge 0.
$$

Proof. $quad$ Note that the sequence $(c_m)$ defined by $c_m=sum_k=0^mbinomklbinomm-kq$ is the convolution of $(a_k)$ and $(b_k),$ where $a_k=binomkl$ and $b_k=binomkq.$ Thus, $c_m$ is the coefficient of $z^m$ in $A(z)B(z),$ where $A(z)$ and $B(z)$ are the generating functions of $(a_k)$ and $(b_k),$ respectively. But, $A(z)=z^l/(1-z)^l+1$ and $B(z)=z^q/(1-z)^q+1,$ so that
$$
A(z)B(z) = fracz^l+q(1-z)^l+q+2=frac1zfracz^l+q+1(1-z)^l+q+2= frac1zsum_kge0binomkl+q+1z^k.
$$
Finally,
$$
c_m=[z^m]A(z)B(z)= binomm+1l+q+1. tag*$blacksquare$
$$

Setting $m:=2n, l:=1,$ and $q:= n-1,$ we get:
$$
sum_k=0^2nbinomk1binom2n-kn-1 = binom2n+1n+1,
$$
and it is easy to see that $sum_k=0^2nbinomk1binom2n-kn-1 = sum_k=1^n+1kbinom2n-kn-1.$