I am trying to turn the recurrence relation into a linear equation in order to sum some probabilities. I've done this with some recursive stuff similar to the Fibonacci sequence, but I was told by a friend that this is much harder because I have an alternating variable.
$x_n=(1-x_{n-1})*(\frac{1}{36+12(-1)^{n}}),$
$x_1=\frac{1}{24}, x_2=\frac{23}{1152}, x_3$ should equal $\frac{1129}{27648}$
The recursive sequence works as intended, and I could definitely program it to approach the eventual number I want, but I'd rather get the proper linear one. I tried to apply the method some people on here explained to me for fibonacci-esque sequences, but I couldn't get it to work. Honestly I don't think I completely get what exactly is happening when I translate from one to the other, so that might be even more important than this specific equation. If I did not explain anything right please ask me, thank you.
Alternate terms of your sequence satisfy linear recurrences with constant coefficients: \begin{align} x_{2n+1}&=\frac{1}{24}\big(1-x_{2n}\big)\\ &=\frac{1}{24}-\frac{1}{24\cdot48}\big(1-x_{2n-1}\big)\\ &=\frac{47}{1152}+\frac{x_{2(n-1)+1}}{1152}\ . \end{align} The solution of this recurrence is \begin{align} x_{2n+1}&=\frac{x_1}{1152^n}+47\left(\frac{1152^n-1}{1152^{n+1}-1152^n}\right)\\ &=\frac{1}{24\cdot1152^n}+\frac{47\big(1152^n-1\big)}{1151\cdot1152^n}\\ &=\frac{1128\cdot1152^n+23}{27624\cdot1152^n}\ , \end{align} and the evenly indexed terms of the sequence are then given by \begin{align} x_{2n}&=1-24x_{2n+1}\\ &=\frac{23\big(1152^n-1\big)}{1151\cdot1152^n}\ . \end{align}