How prove this identity$\sum_{k=0}^{n}\binom{2k}{k}\binom{n+k}{2k}(s-t)^{n-k}t^k=\sum_{k=0}^{n}\binom{n}{k}^2s^{n-k}t^k$

Question

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

Answer 1

$$\begin{align} &\sum_{k=0}^{n}\color{green}{\binom{2k}k\binom {n+k}{2k}}(s-t)^{n-k}t^k\\ &=\sum_{k=0}^{n}\color{green}{\binom {n+k}{2k}\binom{2k}k}\color{blue}{(s-t)^{n-k}}t^k\\ &=\sum_{k=0}^{n}\color{green}{\binom {n+k}{k}\binom nk} t^k\color{blue}{\sum_{r=0}^{n-k}\binom{n-k}rs^{n-k-r}(-t)^r}\\ &=\color{orange}{\sum_{k=0}^{n}\sum_{r=0}^{n-k}}\binom{n+k}k\binom nk\binom{n-k}rs^{n-k-r}t^{k+r}(-1)^r\\ &=\color{orange}{\sum_{j=0}^{n}\sum_{k=0}^{j}}\binom{n+k}k\color{purple}{\binom nk\binom{n-k}{j-k}}s^{n-j}t^{j}(-1)^{j-k}\\ &=\sum_{j=0}^{n}s^{n-j}t^j\sum_{k=0}^{j}\color{red}{\binom{n+k}k}\color{purple}{\binom n{n-k}\binom{n-k}{n-j}}(-1)^{j-k}\\ &=\sum_{j=0}^{n}s^{n-j}t^j\sum_{k=0}^{j}\color{red}{\binom{-n-1}k (-1)^k}\color{purple}{\binom n{n-j}\binom{j}{k}}(-1)^{j-k}\\ &=\sum_{j=0}^{n}s^{n-j}t^j\sum_{k=0}^{j}\binom{-n-1}k\color{purple}{\binom nj\binom j{j-k}}(-1)^{j}\\ &=\sum_{j=0}^{n}\binom nj(-1)^{j}s^{n-j}t^j\color{darkgreen}{\sum_{k=0}^{j}\binom{-n-1}k\binom j{j-k}}\\ &=\sum_{j=0}^{n}\binom nj(-1)^{j}s^{n-j}t^j\color{darkgreen}{\binom{-n-1+j}j}\\ &=\sum_{j=0}^{n}\binom nj(-1)^{j}s^{n-j}t^j\color{darkgreen}{\binom nj (-1)^j}\\ &=\sum_{j=0}^{n}\color{darkred}{\binom nj}s^{n-j}t^j\color{darkred}{\binom nj}\\ &=\sum_{j=0}^{n}\color{darkred}{{\binom nj}^2} s^{n-j}t^j\\ &=\sum_{k=0}^{n}{\binom nk}^2 s^{n-k}t^k\qquad \blacksquare \end{align}$$

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NB: The above also shows that

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

Answer 2

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.