It is well-known that \begin{align*} \sum_{i=0}^n \binom{2i}{i}\binom{2n-2i}{n-i} = 4^n, \end{align*} where one might use combinatorial arguments or generating function technique to prove this.

Now I am interested in finding a closed-form formula for \begin{align*} \sum_{i=0}^n \binom{2i}{i}\binom{2n-2i}{n-i}x^i. \end{align*}

I had a look around but there are not so many literatures that mention this. I came across a paper, and in equation $(5)$ the author did talk about this, but it just doesn't look simple enough to me.

Does anyone have any idea how to proceed? Any idea would be very much appreciated.


2 Answers 2


If we define $$R_n(x)=\sum_{i=0}^{n} {2i\choose i}{2n-2i\choose n-i}x^i$$ then the generating function for these polynomials can be calculated:

$$G(x,t)=\sum_{n=0}^{\infty}R_n(x)t^n\\=\sum_{i=0}^{\infty}{2i\choose i}(xt)^i\sum_{n=0}^{\infty}{2n\choose n}t^n\\=\frac{1}{\sqrt{(1-4t)(1-4xt)}}$$

(we can combine the square roots in their common region of convergence) However we note that


and using the expression for the generating function of the Legendre polynomial $P_n(x)$ and it's uniqueness we may deduce that:

$$R_n(x)=4^n x^{n/2} P_n\Big(\frac{x+1}{2\sqrt{x}}\Big)$$

The generating function approach is in general very useful for calculating quantities like $R_n^{(m)}(1)$ as well.

For example, a simple calculation shows that

$$R_n^{'}(1)=\frac{\partial}{\partial x}G(e^x,t)\Big|_{x=0}=\frac{1}{n!}\frac{d^{n}}{dt^n}\Big(\frac{2t}{(1-4t)^2}\Big)\Big|_{t=0}=n2^{2n-1}$$

  • 1
    $\begingroup$ This is beautiful, for sure. $\to +1$ $\endgroup$ Commented May 10, 2019 at 8:35

I think that you are facing a gaussian hypergeometric function (this hides an infinite summation) $$\sum_{i=0}^n \binom{2i}{i}\binom{2n-2i}{n-i}x^i=\binom{2 n}{n} \, _2F_1\left(\frac{1}{2},-n;-n+\frac{1}{2};x\right)$$


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