Two weeks ago I asked this question on here to figure out a way to prove Rodrigues' formula for Gegenbauer Polynomials without going via Jacobi Polynomials or the orthogonality criterion. I didn't actually find a way to do this, so I tried to instead prove orthogonality, only based on the generating function, without Rodrigues. My proof used the formula $$C_n^\lambda(\cos(\theta))=\sum_{k=0}^n\frac{(\lambda)_k(\lambda)_{n-k}}{k!(n-k)!}\cos((n-2k)\theta),$$ where $(\lambda)_k=\lambda(\lambda+1)\cdots (\lambda+k-1)$ is the rising factorial. With this I find \begin{equation} \begin{split} \int_{-1}^1 (C_n^\lambda(s))^2\,(1-s^2)^{\lambda-\frac{1}{2}}\,\mathrm{d} s= \sum_{j,k=0}^n&\frac{(\lambda)_k(\lambda)_{n-k}}{k!(n-k)!}\cdot \frac{(\lambda)_j(\lambda)_{n-j}}{j!(n-j)!}\\ &\times \int_0^\pi\cos((n-2j)\theta)\cos((n-2k)\theta)\sin^{2\lambda}(\theta)\,\mathrm{d}\theta. \end{split} \end{equation} From here I could use the fact that $\cos((n-2j)\theta)\cos((n-2k)\theta)=\frac{1}{2}\left(\cos(2(n-j-k)\theta)+\cos(2(j-k)\theta)\right),$ thus obtaining two integrals, which can actually be evaluated. Some further computation got me to the following result $$\int_{-1}^1 (C_n^\lambda(s))^2\,(1-s^2)^{\lambda-\frac{1}{2}}\,\mathrm{d} s=\frac{\pi 2^{1-2\lambda}\Gamma(2\lambda)}{\Gamma(\lambda)^2}\sum_{j,k=0}^n\frac{(\lambda)_k(\lambda)_{n-k}(\lambda)_j(\lambda)_{n-j}}{k!(n-k)!j!(n-j)!}\frac{(-\lambda)_{|j-k|}}{(\lambda)_{|j-k|+1}}$$ I know what the result is supposed to be (since I'd know how to apply Rodrigues, if I had Rodrigues), so I get the following equation that I want to prove: $$\frac{(2\lambda)_n}{n!}\frac{1}{n+\lambda}=\sum_{j,k=0}^n\frac{(\lambda)_k(\lambda)_{n-k}(\lambda)_j(\lambda)_{n-j}}{k!(n-k)!j!(n-j)!}\frac{(-\lambda)_{|j-k|}}{(\lambda)_{|j-k|+1}}$$ Multiplying be $(n!)^2\cdot (\lambda)_{n+1}$ yields a polynomial in $\lambda$ with integer coefficients, as follows: $$n!(2\lambda)_n(\lambda)_n=\sum_{j,k=0}^n\binom{n}{k}\binom{n}{j}(\lambda)_k(\lambda)_{n-k}(\lambda)_j(\lambda)_{n-j}(-\lambda)_{|j-k|}(\lambda+|j-k|+1)_{n-|j-k|}.$$ This is where I am stuck. This very much looks like some combinatorial argument, in particular if we realise that $\lambda\in\mathbb{N}$ would give us only integers. I also evaluated the sum for $n=0,1,2,3$ and got the correct result, then tried to do some comparing of coefficients, which also worked fine, so I am very confident that there were no mistakes here. Yet I really don't see a way to actually prove this statement, so maybe the combinatorial minds (or the analytical minds, I guess) of stackexchange have some ideas how to continue. Maybe there is something I overlooked when trying to prove the statement on the natural numbers and then generalising by analytic continuation? I much appreciate any ideas!
The following formulas might be useful, but I haven't been able to use them very well so far: $$\sum_{k=0}^n\frac{(\lambda)_k(\lambda)_{n-k}}{k!(n-k)!}=\frac{(2\lambda)_n}{n!},\qquad \sum_{k=0}^{2n}\frac{(\lambda)_k(\lambda)_{2n-k}}{k!(2n-k)!}(-1)^{n-k}=(-1)^n\frac{(\lambda)_{n}}{n!}\quad \text{and} \quad \sum_{k=0}^{2n+1}\frac{(\lambda)_k(\lambda)_{2n+1-k}}{k!(2n+1-k)!}\sin((n-k)\pi)=0$$