Consider a birth-death process with parameters $\lambda_n = 2 \cdot (6-n)$ and $\mu_n = 3.5 \cdot n$ where $n = 0,1,2,\dots,6$. Find the stationary distribution (as a function of n).
My attempt is that by using the balance equation we get $2(6-n)\pi_n = 3.5(n+1)\pi_{n+1} \\ \frac{2(6-n)}{3.5n}\pi_n = \pi_{n+1} \\ \pi_n = \frac{2(7-n)}{3.5(n-1)}\cdot\frac{2(8-n)}{3.5(n-2)}\cdot\frac{2(9-n)}{3.5(n-3)}\dots\\ \pi_n = \frac{2^n\cdot(7-n)(8-n)(9-n)\dots}{3.5^n\cdot(n-1)!}\cdot\pi_0$
And calculated that $\pi_0 = \frac{117649}{1771561}$ by manually calculating each $\pi_n$ but I don't know how to arrive at a general formula for the part before $\pi_0$ in the $\pi_n$ equation
I would appreciate if anyone can help me arrive at the correct answer