Fdtxjr

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

Starting from

$$nsum_k=1^n+1 2nchoose k^-1 nchoose k-1$$

we find

$$nsum_k=1^n+1 fracn!(k-1)! times (n+1-k)!
frack! times (2n-k)!(2n)!
\ = fracntimes n!(2n)!
sum_k=1^n+1 frack(n+1-k)! (2n-k)!
\ = fracn!^2(2n)!
sum_k=1^n+1 k 2n-kchoose n+1-k.$$

This is

$$fracn!^2(2n)!
sum_k=0^n+1 k [z^n+1-k] (1+z)^2n-k
\ = fracn!^2(2n)!
[z^n+1] (1+z)^2n
sum_k=0^n+1 k z^k (1+z)^-k$$

Now when $kgt n+1$ there is no contribution to the coefficient
extractor and we get

$$fracn!^2(2n)!
[z^n+1] (1+z)^2n
sum_kge 0 k z^k (1+z)^-k
\ = fracn!^2(2n)!
[z^n+1] (1+z)^2n
fracz/(1+z)(1-z/(1+z))^2
\ = fracn!^2(2n)!
[z^n+1] (1+z)^2n+2 fracz1+z
\ = fracn!^2(2n)!
[z^n] (1+z)^2n+1
= fracn!^2(2n)! 2n+1choose n
\ = n!^2 times
(2n+1) times frac1n! times (n+1)!
= frac2n+1n+1.$$

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.

$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
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newcommandverts[1]leftvert,#1,rightvert$
beginalign
sum_k = 1^n + 1nn choose k - 1 over
2n choose k & =
nsum_k = 0^nn choose k over 2n choose k + 1
\[5mm] & =
nsum_k = 0^nn choose k
1 over
pars2n!/bracksparsk + 1!pars2n - k - 1!
\[5mm] & =
npars2n + 1sum_k = 0^n
n choose kGammaparsk + 2Gammapars2n - k! over Gammapars2n + 2
\[5mm] & =
npars2n + 1sum_k = 0^n
n choose kint_0^1
t^k + 1pars1 - t^2n - k - 1,dd t
\[5mm] & =
npars2n + 1int_0^1pars1 - t^2n,
t over 1 - tsum_k = 0^nn choose k
parst over 1 - t^k,dd t
\[5mm] & =
npars2n + 1int_0^1pars1 - t^2n - 1, t
pars1 + t over 1 - t^n,dd t
\[5mm] & =
npars2n + 1int_0^1t, pars1 - t^n - 1,dd t \[5mm] & =
npars2n + 1int_0^1pars1 - t,t^n - 1,dd t
\[5mm] & =
npars2n + 1pars1 over n - 1 over n +1
= bbx2n + 1 over n + 1
endalign

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 $sum_k=0^2nbinomk1binom2n-kn-1 = sum_k=1^n+1kbinom2n-kn-1.$