Hi I am having trouble on how to solve for the odd terms of recurrence relation in terms of exponential and factorials. How are you able to see a pattern to simplify a non standard factorial.
This is the recurrence relation $A_{n+2}(n+2)(n+1) + 4A_n(n(n-1)-8)$
$A_{n+2} = -4A_n(n-2)/(n+2)$
$A_{2n}=0$ for n greater than or equal to 2
Odd terms what I have so far $A_{2n+1} = (-1)^{n+1}*4^n /(3)(5)(7)...(2n+1)$
How can you see that denominator is simplifed to $(2n-1)(2n+1)$
The answer to the question in the book is $A_n = (-1)^{n+1}*4^n/(2n-1)(2n+1)$
Check your work for the terms with odd index: for odd $n$ you get
$$\begin{align*} A_n&=-4\cdot\frac{n-4}n\cdot A_{n-2}\\ &=(-4)^2\cdot\frac{n-4}n\cdot\frac{n-6}{n-2}\cdot A_{n-4}\\ &=(-4)^3\cdot\frac{\color{red}{n-4}}n\cdot\frac{n-6}{n-2}\cdot\frac{n-8}{\color{red}{n-4}}\cdot A_{n-6}\\ &=(-4)^4\cdot\frac{\color{red}{n-4}}n\cdot\frac{\color{blue}{n-6}}{n-2}\cdot\frac{n-8}{\color{red}{n-4}}\cdot\frac{n-10}{\color{blue}{n-6}}\cdot A_{n-8}\\ &=\ldots\;, \end{align*}$$
and all of the factors $n-2k$ in the numerator are cancelled out by factors in the denominator. Only the first two factors in the denominator remain.