# Going from recurrence relations to closed form

How do I go from the following recurrence relation

a(n) = (n+1)a(n-1) where a(0) = 2


to a closed form? I know I need to use an iterative approach but I am not quite sure what to look for.

We can divide through by the summation factor $(n+1)!$ to get $${a_n\over (n+1)!}=\frac{a_{n-1}}{n!}$$ Now let $$S_n:=\frac{a_n}{(n+1)!}$$ Thus $S_n=S_{n-1}=c$ is constant, but $a_0=2$ and so $S_0=2/1=2\implies S_n\equiv2$ Thus we have $$a_n=(n+1)!S_n=2\cdot (n+1)!$$
• I'm not sure - I learned it originally from the book "Concrete Mathematics" by Knuth, et al. See this: homepages.gac.edu/~holte/courses/mcs256/documents/… They show how to find $S_n$ in general. This method can be used to solve any recurrence of the form $a_nx_n=b_nx_{n-1}+c_n$ so it's fairly powerful and nice to know. – user142299 Apr 28 '14 at 1:34