is there a specified way to get this expansion?

Question

$\beta(x,y-1)=\beta(x,y)+\beta(x+1,y)+\beta(x+2,y)........$ i have tried to find the proof of this formula on google and some books but i found other methods like this MSE post Beta function series expansion and can not understand them can anyone help me and give me a simple method to prove it via gamma function or another method

Answer 1

\begin{align*} \beta(x,y)+&\beta(x+1,y)+\beta(x+2,y)+\cdots\\ &= \beta(x,y)+\frac{x}{(x+y)}\beta(x,y)+\frac{x(x+1)}{(x+y)(x+y+1)}\beta(x,y)+\cdots\\ &=\beta(x,y)\left[1+\frac{x}{x+y}+\frac{x(x+1)}{(x+y)(x+y+1)}+\cdots\right]\\ &=\beta(x,y)\sum_{k=0}^\infty\frac{(x)_k}{(x+y)_k},\qquad (x)_k=x(x+1)\cdots(x+k-1)\\ &=\beta(x,y)\sum_{k=0}^\infty\frac{(1)_k(x)_k}{k! (x+y)_k},\qquad (1)_k=k!\\ &=\beta(x,y) {}_2F_1(1,x;x+y;1)\\ &=\beta(x,y)\frac{\Gamma(x+y)\Gamma(y-1)}{\Gamma(x+y-1)\Gamma(y)},\quad using~Gauss’~summation~ formula\\ &=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\frac{\Gamma(x+y)\Gamma(y-1)}{\Gamma(x+y-1)\Gamma(y)},\qquad for ~\beta(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\\ &=\frac{\Gamma(x)\Gamma(y-1)}{\Gamma(x+y-1)}=\beta(x,y-1). \end{align*}