I found the Christoffel–Darboux formula of the Laguerre polynomials on wikipedia, https://en.wikipedia.org/wiki/Laguerre_polynomials
\begin{align} K_n^{(\alpha)}(x,y) &:= \frac{1}{\Gamma(\alpha+1)} \sum_{i=0}^n \frac{L_i^{(\alpha)}(x) L_i^{(\alpha)}(y)}{{\alpha+i \choose i}}\ &{=}\frac{1}{\Gamma(\alpha+1)} \frac{L_n^{(\alpha)}(x) L_{n+1}^{(\alpha)}(y) - L_{n+1}^{(\alpha)}(x) L_n^{(\alpha)}(y)}{\frac{x-y}{n+1} {n+\alpha \choose n}} \ &{=}\frac{1}{\Gamma(\alpha+1)}\sum_{i=0}^n \frac{x^i}{i!} \frac{L_{n-i}^{(\alpha+i)}(x) L_{n-i}^{(\alpha+i+1)}(y)}{{\alpha+n \choose n}{n \choose i}}; \end{align}
but there is no reference for the proof of the last equation. I have tried for several days, but failed.
I used the definition of Laguerre polynomials
\begin{equation} L_n^{(\alpha)} (x) = \sum_{i=0}^n (-1)^i {n+\alpha \choose n-i} \frac{x^i}{i!} \end{equation}
and the expansions
\begin{equation}\frac{x^n}{n!}= \sum_{i=0}^n (-1)^i {n+ \alpha \choose n-i} L_i^{(\alpha)}(x),\end{equation}
\begin{equation}L_n^{(\alpha)}(x)= \sum_{i=0}^n L_{n-i}^{(\alpha+i)}(y)\frac{(y-x)^i}{i!},\end{equation}
\begin{equation}L_n^{(\alpha+1)}(x)= \sum_{i=0}^n L_i^{(\alpha)}(x)\end{equation}
The generating function for Laguerre polynomials is
\begin{equation}\sum_n^\infty t^n L^{(\alpha)}_n(x)= \frac{1}{(1-t)^{\alpha+1}} e^{-\frac{tx}{1-t}}.\end{equation}
Some other references are listed as follows. I am not sure if they would help.
[2] doi.org/10.1016/j.jat.2009.07.006
Can anyone help me on the proof of it?
After some work on the expansions to compare the coefficients of $x^jy^k$, I came to the formulas as follows
(a) \begin{equation}\sum_{i=\max\{j,k\}}^{n}{i\choose j}{i+\alpha \choose i-k}\end{equation}
(b) \begin{equation} \sum_{i_1=\max\{n-j,k\}}^n (-1)^{-n+i_1} {i_1 \choose n-j}{n+\alpha+1 \choose i_1-k} \end{equation} where $j=0,1,2...,n;k=0,1,2,...,n$ and I need now to proof (a)=(b)
