If $x+y=1$ then $x^{n+1}\sum\limits_{k=0}^n \binom{n+k}{k}y^k+y^{n+1}\sum\limits_{k=0}^n \binom{n+k}{k}x^k=1$ $$x^{n+1}\sum_{k=0}^n \binom{n+k}{k}y^k+y^{n+1}\sum_{k=0}^n \binom{n+k}{k}x^k=1$$
with one property such that x+y=1
I tried using the binomial equation to replace $y^k$ with $(1-x)^k$ but that will not cancel out anything! so I tried to use 
$$\sum_{k=0}^n \binom{n}{k} x^k=(1+x)^n$$
which when $x=1$ gives the following
$$\sum_{k=0}^n \binom nk=2^n\\
\sum_{k=1}^n\binom nk=2^n-1$$
so I hope to combine the two such that $x+y=1$ and I can use the above equation? thanks

I just copy my answer of this question.

$$\forall m,n\in\Bbb N,\ \forall x\in\Bbb R,\ \sum_{k=0}^{m}\binom{n+k}{k}(1-x)^kx^{n+1}=1-\sum_{k=0}^{n}\binom{m+k}{k}x^k(1-x)^{m+1}$$

Let $m,n\in\Bbb N$, $f:x\mapsto\sum_{k=0}^{m}\binom{n+k}{k}(1-x)^kx^{n+1}$, we have $f(0)=0$ and for all $x\in\Bbb R$,
\begin{align}
f'(x)&=(n+1)x^{n}\sum_{k=0}^{m}\binom{n+k}{k}(1-x)^k-x^{n+1}\sum_{k=0}^{m}\binom{n+k}{k}k(1-x)^{k-1}\\
&=x^{n}(1-x)^{m}\cdot A
\end{align}
where
\begin{align}
A&=\sum_{k=0}^{m}\binom{n+k}{k}(n+1)(1-x)^{k-m}-\sum_{k=0}^{m}\binom{n+k}{k}kx(1-x)^{k-m-1}\\
&=\sum_{k=0}^{m}\binom{n+k}{k}(n+k+1)(1-x)^{k-m}-\sum_{k=0}^{m}\binom{n+k}{k}k(1-x)^{k-m-1}\tag1\\
&=\sum_{k=1}^{m+1}\binom{n+k}{k}k(1-x)^{k-m-1}-\sum_{k=1}^{m}\binom{n+k}{k}k(1-x)^{k-m-1}\tag2\\
&=\binom{n+m+1}{m+1}(m+1)\\
&=\frac{(n+m+1)!}{n!\ m!}
\end{align}
$(1):-x(1-x)^{k-m-1}=(1-\frac1{1-x})(1-x)^{k-m}=(1-x)^{k-m}-(1-x)^{k-m-1}$
$(2):\binom{n+k}{k}(n+k+1)=\frac{(n+k+1)!}{n!k!}=\binom{n+k+1}{k+1}(k+1)$
Thus,
\begin{align}
\forall x\in \Bbb R,\ f(x)&=A\int_0^xt^n(1-t)^mdt\\
&=A\int_{1-x}^1(1-s)^ns^mds\tag{$s=1-t$}\\
&=A\int_0^1(1-s)^ns^mds-A\int_0^{1-x}(1-s)^ns^mds\\
&=1-\sum_{k=0}^{n}\binom{m+k}{k}x^k(1-x)^{m+1}\tag3
\end{align}
$(3):$ integration by parts proves that $\int_0^1(1-s)^ns^mds=\frac{n!\ m!}{(n+m+1)!}=A^{-1}$, and $m\leftrightarrow n$ and $x\leftrightarrow 1-x$ in the expressions of $f$ give the second term.
