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$$x^{n+1}\sum_{k=0}^n \binom{n+k}{k}y^k+y^{n+1}\sum_{k=0}^n \binom{n+k}{k}x^k=1$$

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

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  • $\begingroup$ From the first equation, multiply $x^n$ and $y^n$ back in the sums. Then using binomial formula you will end up with $(x+y)^{n+k}=1$. Then you are left with $x+y$. $\endgroup$ – Harry Feb 7 '18 at 8:17
  • $\begingroup$ how does $$x^n*y^k$$ with the summation of n+k choose n end up being $$(x+y)^{n+k}=1$$ $\endgroup$ – james black Feb 7 '18 at 8:51
  • $\begingroup$ It’s from the Binomial formula $(x+y)^{n+k}$ is equal to the sum of $(n+k)C(k)x^{k}y^{n}$. And here I assume you are given $(x+y)=1$. My only worry is about the end point of your sum and the Binomial coefficients do not match. $\endgroup$ – Harry Feb 7 '18 at 9:13
  • $\begingroup$ @Harry Offtopic ($k$ is a running index). $\endgroup$ – Did Feb 7 '18 at 10:13
  • $\begingroup$ This looks like Gosper's Identity. I posted a similar question on it here some time ago. There are some good answers there. $\endgroup$ – hypergeometric Feb 7 '18 at 15:23
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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.

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