Proving a combinatorial identity related to convolution of central binomial coefficients

I'm trying to calculate the coefficient values for $$f(x) = \sum_{k=0}^n {n \choose k}^2 (1+x)^{2n-2k}(1-x)^{2k}$$. TL;DR I don't know where to begin in order to prove:

$$\sum_{k=0}^r (-4)^k{n \choose k}{2n-2k \choose n}{n-2k \choose r-k} = {2r \choose r}{2n-2r \choose n-r}$$

Explanation: writing $$f(x)$$ as the coefficient of $$y^n$$ in:

\begin{align*} g(x, y) & = [y + (1 + x)^2]^n[y - (1 + x)^2]^n \\\\ & = [(y + 1 + x^2)^2 - 4x^2]^n \end{align*}

shows that this is an even function in $$x$$, so any nonzero coefficient of $$x^l$$ requires $$l = 2r$$ for some integer $$r$$. Making this substitution and calculating the coefficient of $$y^n$$ using the Multinomial Theorem gives

\begin{align*} & [x^{2r}][y^n]\left((y+1+x^2)^2 - 4x^2 \right)^n \\ = & [x^{2r}]\sum_{k=0}^n {2n-2k \choose k; n-2k} (-4)^{k}x^{2k}(1+x^2)^{n-2k} \;\text{(Multinomial Expansion)} \\ = & [x^{r}]\sum_{k=0}^n (-4)^k {n \choose k}{2n-2k \choose n}x^k(1+x)^{n-2k} \;\text{(x^2 \to x substitution)} \\ = & \sum_{k=0}^r (-4)^k{n \choose k}{2n-2k \choose n}{n-2k \choose r-k} \end{align*}

At this point, I can tell by numerical calculation of the first few values of $$n$$ and $$r$$ that this is equal to $${2r \choose r}{2n-2r \choose n-r}$$, but despite many attempts at resolving the sum using generating functions and combinatorial arguments I haven't managed to prove it. I would appreciate any help with this or insight into my methods, the sum, or how to prove the value of the coefficient.

Suppose we seek to evaluate the coefficients

$$[x^q] f_n(x) = [x^q] \sum_{k=0}^n {n\choose k}^2 (1+x)^{2n-2k} (1-x)^{2k}.$$

Recall the generating function of the Legendre polynomials

$$\sum_{n\ge 0} P_n(y) t^n = \frac{1}{\sqrt{1-2yt+t^2}}$$

and the known fact

$$P_n(y) = \left[\frac{y-1}{2}\right]^n \sum_{k=0}^n {n\choose k}^2 \left[\frac{y+1}{y-1}\right]^k.$$

Now put $$y= -\frac{x^2+1}{2x}$$ to get

$$P_n\left(-\frac{x^2+1}{2x}\right) = (-1)^n \frac{(1+x)^{2n}}{4^n x^n} \sum_{k=0}^n {n\choose k}^2 \left[\frac{(1-x)^2}{(1+x)^2}\right]^k.$$

It follows that

$$f_n(x) = (-1)^n 4^n x^n P_n\left(-\frac{x^2+1}{2x}\right).$$

Using the OGF we have

$$(-1)^n 4^n x^n [t^n] \frac{1}{\sqrt{1+((x^2+1)/x)t+t^2}} \\ = [t^n] \frac{1}{\sqrt{1-4(x^2+1)t+16x^2t^2}} \\ = [t^n] \frac{1}{\sqrt{1-4t}} \frac{1}{\sqrt{1-4x^2t}}.$$

We observe at this point that we must have $$q=2r$$ and continue with

$$[x^{2r}] [t^n] \frac{1}{\sqrt{1-4t}} \frac{1}{\sqrt{1-4x^2t}} \\ = [x^r] [t^n] \frac{1}{\sqrt{1-4xt}} \frac{1}{\sqrt{1-4t}} = [x^r] \sum_{k=0}^n {2k\choose k} x^k {2n-2k\choose n-k} \\ = {2r\choose r} {2n-2r\choose n-r}.$$

This is the claim. One reference for Legendre polynomials is Gould's Combinatorial Identities (page 38).

• This is perfect -- I had worked backward from the result to get $$[t^n] \frac{1}{\sqrt{1-4t}} \frac{1}{\sqrt{1-4x^2t}}$$ but I couldn't figure out how to bridge the gap since I've never used the generating function of the Legendre Polynomials before. Thanks for the help, and the reference! Apr 28, 2022 at 23:39
• Thank you for the kind remark. Good to know it helped. Apr 29, 2022 at 0:20