I am trying to find a solution the product series, but denominator has a form x+y, so it makes complicated calculation as follows. Please help to find a product.
$\prod_{i=1}^{k-1}\frac{2a(n-i)}{(n-i)^2(n-i+1)+2b}$
I really appreciated your help. Thank you.
I do not see what you could do beside using Pochhammer symbols.
Expand the denominator $$(n-i)^2(n-i+1)+2b=\left(2 b+n^2+n^3\right)- n\left(3 n+2 \right)i+ (3 n+1)i^2-i^3$$ and let $(r,s,t)$ to be the roots of the cubic in $i$. So $$P_k=\prod_{i=1}^{k-1}\frac{2a(n-i)}{(n-i)^2(n-i+1)+2b}=\frac{\prod_{i=1}^{k-1}2a(i-n)} {\prod_{i=1}^{k-1} (i-r) \prod_{i=1}^{k-1} (i-s)\prod_{i=1}^{k-1} (i-t)}$$ $$P_k=\frac{(2a)^{k-1} (1-n)_{k-1}}{(1-r)_{k-1} (1-s)_{k-1} (1-t)_{k-1}}$$
Assuming $b >0$, the cubic equation has only one real root $$r=n+\frac{1}{3} \left(1-2 \cosh \left(\frac{1}{3} \cosh ^{-1}(1+27 b)\right)\right)$$ and two complex conjugate roots which can write $s=c+I\,d$ and $t=c-I\,d$ (first time I have to write a capital $I$ for $I^2=-1$); in fact, by Vieta, we know the sum $(r+s+t)$ and the product $(r\,s\,t)$ so we know $c$ and $d$. $$\prod_{i=1}^{k-1} (i-s)\prod_{i=1}^{k-1} (i-t)=(1-c-I d)_{k-1} (1-c+I d)_{k-1}$$