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Today I see a paper,and this author say it is easy to have this identity.But I take sometimes to prove it,and I can't prove it.

show this following identity holds for any real $s$ and $t$ and any postive integer $n$ $$\sum_{k=0}^{n}\binom{2k}{k}\binom{n+k}{2k}(s-t)^{n-k}t^k=\sum_{k=0}^{n}\binom{n}{k}^2s^{n-k}t^k$$

My some idea: $$\binom{2k}{k}\binom{n+k}{2k}=\dfrac{(2k)!}{(k!)^2}\cdot\dfrac{(n+k)!}{(2k)!(n-k)!}=\dfrac{n!}{k!(n-k)!}\cdot\dfrac{(n+k)!}{n!k!}=\binom{n}{k}\binom{n+k}{k}$$ so we must show this following identity $$\sum_{k=0}^{n}\binom{n}{k}\binom{n+k}{k}(s-t)^{n-k}t^k=\sum_{k=0}^{n}\binom{n}{k}^2s^{n-k}t^k$$ then I can't works

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  • $\begingroup$ Why not work out $(s-t)^{n-k}$ by the binomial formula, and the compare coefficients of each monomial? $\endgroup$ – Marc van Leeuwen Dec 12 '14 at 8:13
  • $\begingroup$ I have try it, I know you meaning $(s-t)^{n-k}=\sum_{i=0}^{n-k}\binom{n-k}{i}s^i(-t)^{n-k-i}$,then it is very ugly $\endgroup$ – math110 Dec 12 '14 at 8:15
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$$\begin{align} &\sum_{k=0}^{n}\color{green}{\binom{2k}k\binom {n+k}{2k}}(s-t)^{n-k}t^k\\ &=\sum_{k=0}^{n}\color{green}{\binom {n+k}{2k}\binom{2k}k}\color{blue}{(s-t)^{n-k}}t^k\\ &=\sum_{k=0}^{n}\color{green}{\binom {n+k}{k}\binom nk} t^k\color{blue}{\sum_{r=0}^{n-k}\binom{n-k}rs^{n-k-r}(-t)^r}\\ &=\color{orange}{\sum_{k=0}^{n}\sum_{r=0}^{n-k}}\binom{n+k}k\binom nk\binom{n-k}rs^{n-k-r}t^{k+r}(-1)^r\\ &=\color{orange}{\sum_{j=0}^{n}\sum_{k=0}^{j}}\binom{n+k}k\color{purple}{\binom nk\binom{n-k}{j-k}}s^{n-j}t^{j}(-1)^{j-k}\\ &=\sum_{j=0}^{n}s^{n-j}t^j\sum_{k=0}^{j}\color{red}{\binom{n+k}k}\color{purple}{\binom n{n-k}\binom{n-k}{n-j}}(-1)^{j-k}\\ &=\sum_{j=0}^{n}s^{n-j}t^j\sum_{k=0}^{j}\color{red}{\binom{-n-1}k (-1)^k}\color{purple}{\binom n{n-j}\binom{j}{k}}(-1)^{j-k}\\ &=\sum_{j=0}^{n}s^{n-j}t^j\sum_{k=0}^{j}\binom{-n-1}k\color{purple}{\binom nj\binom j{j-k}}(-1)^{j}\\ &=\sum_{j=0}^{n}\binom nj(-1)^{j}s^{n-j}t^j\color{darkgreen}{\sum_{k=0}^{j}\binom{-n-1}k\binom j{j-k}}\\ &=\sum_{j=0}^{n}\binom nj(-1)^{j}s^{n-j}t^j\color{darkgreen}{\binom{-n-1+j}j}\\ &=\sum_{j=0}^{n}\binom nj(-1)^{j}s^{n-j}t^j\color{darkgreen}{\binom nj (-1)^j}\\ &=\sum_{j=0}^{n}\color{darkred}{\binom nj}s^{n-j}t^j\color{darkred}{\binom nj}\\ &=\sum_{j=0}^{n}\color{darkred}{{\binom nj}^2} s^{n-j}t^j\\ &=\sum_{k=0}^{n}{\binom nk}^2 s^{n-k}t^k\qquad \blacksquare \end{align}$$


NB: The above also shows that

$$\small\begin{align} &\sum_{k=0}^{j}\underbrace{\binom{n+k}k\binom nk\binom{n-k}{j-k}\color{gray}{\binom{n-j}{n-j}}}_{\Large\binom{n+k}{k,k,j-k,n-j}}(-1)^{j-k}\\ &=\sum_{k=0}^{j}\binom{n+k}{k,k,j-k,n-j}(-1)^{j-k}\\ &={\binom nj}^2 \end{align}$$

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  • $\begingroup$ It's very very nice! but why $$\sum_{k=0}^{j}\binom{n+k}{k,k,j-k,n-j}(-1)^{j-k}=\binom{n}{j}^2?$$ $\endgroup$ – math110 Dec 13 '14 at 11:11
  • $\begingroup$ @math110 - Thank you! That's very kind of you.:) With regard to your question, I've edited the solution for greater clarity. Hope this is helpful! $\endgroup$ – hypergeometric Dec 13 '14 at 17:11
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By way of enrichment here is another algebraic proof using basic complex variables.

For the sum on the LHS which is $$\sum_{k=0}^n {2k\choose k} {n+k\choose 2k} (s-t)^{n-k} t^k$$ or rather $$\sum_{k=0}^n {n\choose k}{n+k\choose k} (s-t)^{n-k} t^k$$

introduce the integral representation $${n+k\choose k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n+k}}{z^{k+1}} \; dz.$$

This yields for the sum $$(s-t)^n \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z} \sum_{k=0}^n {n\choose k} \left(\frac{1+z}{z} \frac{t}{s-t} \right)^k \; dz \\ = (s-t)^n \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z} \left(1+ \frac{1+z}{z} \frac{t}{s-t}\right)^n \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z} \left(s-t + t \frac{1+z}{z}\right)^n \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z} \left(s-t + \frac{t}{z} + t\right)^n \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z} \left(s + \frac{t}{z}\right)^n \; dz.$$

For the sum on the RHS which is $$\sum_{k=0}^n {n\choose k}^2 s^{n-k} t^k$$ introduce the integral representation $${n\choose k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{k+1}} \; dz.$$

This yields for the sum the integral $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z} \sum_{k=0}^n {n\choose k} \frac{1}{z^k} s^{n-k} t^k \; dz$$ or $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z} \left(s+\frac{t}{z}\right)^n \; dz.$$

We have the identical integrals for LHS and RHS, done. We have not made use of the properties of complex integrals here so this computation can also be presented using just algebra of generating functions.

There is another computation in the same spirit at this MSE link.

Apparently this method is due to Egorychev although some of it is probably folklore.

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  • $\begingroup$ Very nice. If we were to approach this without using complex integrals, and note that $$\sum_{k=0}^{n}\binom {n+k}{2k}\binom{2k}k\color{blue}{(s-t)^{n-k}}t^k =\sum_{k=0}^{n}\binom {n+k}{k}\binom nk t^k\color{blue}{\sum_{r=0}^{n-k}\binom{n-k}rs^{n-k-r}(-t)^r}\\ =\sum_{k=0}^{n}\binom nk s^{n-k}t^k\color{red}{\sum_{r=0}^{n-k}\binom{n+k}k\binom{n-k}rs^{-r}(-t)^r}$$ can it be shown that the part in red reduces to $\binom nk$? If so then we have the answer, but it doesn't seem likely as it contains $s$ and $t$... $\endgroup$ – hypergeometric Dec 13 '14 at 1:45
  • $\begingroup$ When you ask Maple it just collapses the binomial to where you started from. Owing to the dependence on $s$ and $t$ you cannot conclude that it is equal to ${n\choose k}.$ There are many choices for the red term that would give the same value of the sum. $\endgroup$ – Marko Riedel Dec 13 '14 at 1:55
  • $\begingroup$ Yes, I realise that it collapses back to the starting point. However, having isolated $\binom nk s^{n-k}t^k$ the only missing part of the product is another $\binom nk$ in order to give ${\binom nk}^2$, as in RHS, so logically one would have thought that the red part should reduce to $\binom nk$... $\endgroup$ – hypergeometric Dec 13 '14 at 2:01

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