Sum of square of terms in binomial expansion

Question

We have the binomial expansion as this $$(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k = \sum_{k=0}^n {n \choose k}x^{k}y^{n-k}$$

Is there a formula for $\sum_{k=0}^n {n \choose k}^2x^{2k}y^{2(n-k)}$, the sum when the terms are squared? If that's impossible, maybe one for when $y^2 = 1 - x^2$ if that makes thing easier? I've been thinking about it but couldn't figure anything out. I know it was asked here in a somewhat similar fashion but there was not conclusive answer. Thank you in advance!

I know that there is a formula which says $$ {2n \choose n} = \sum_{k=0}^n {n \choose k}^2$$ — Lucky Chouhan, Sep 04 '23 at 16:42
The comment of @LuckyChouhan is not coincidental. In your original expression, consider the coefficient of $~x^n~$ in the expansion of $~(x + y)^n \times (x+y)^n.~$ As $~k~$ traverses $~0~$ through $~n~$, you will have $~n+1~$ terms to add, each one of the form $$\left[\binom{n}{k} x^k y^{n-k}\right] \times \left[\binom{n}{k} x^{n-k} y^k\right].$$ — user2661923, Sep 04 '23 at 18:27
Re my previous comment, typo: should be the coefficient of $~\color{red}{x^n y^n}.~$ Also, re my previous comment, this doesn't directly address your question, since you are interested in terms of the form $$\binom{n}{k}^2 x^{2k}y^{n-2k}.~$$ However, it might be a reasonable place to start. — user2661923, Sep 04 '23 at 18:31
Apparently there is an identity that would get rid of the squaring, but you'd still have a product of binomial coefficients, in case that ever helps: Prove that $\sum_{l=0}^{n} \binom{n}{l}^2 (x+y)^{2l} (x-y)^{2(n-l)} = \sum_{l=0}^{n} \binom{2l}{l} \binom{2(n-l)}{n-l} x^{2l}y^{2(n-l)}$ — Bruno B, Sep 04 '23 at 19:04
There also seems to be another identity, tis time in the form of an integral, in this answer: An integral involving Bessel function of the first kind of the Sonine-Gegenbauer sort, namely: $$2\pi \sum_{k=0}^{n}\binom{n}{k}^2 a^{2k} b^{2n-2k} = \int_{0}^{2\pi}(a^2+b^2-2ab\cos\phi)^n,d\phi$$ In case that becomes useful somehow? — Bruno B, Sep 04 '23 at 19:45

Sum of square of terms in binomial expansion

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