How do I solve $a_{n+1}=(n+1)a_n$ by recurrence relations given $a_0=1$? I realize that it equals $n!$ by inspection, but how do I solve it as a first-order non-homogenous recurrence relation (with characteristic polynomials like in linear algebra)? I have tried to solve it like this:
$a_{n+1}=(n+1)a_n$
$a_{n}=(n)a_{n-1}$
But it is not like adding a constant, in which I can just do this and subtract. Dividing is of no help either. How should I approach such a problem?
P. S. I have not learned differential equations, and don't wish to do algebra/function bash as in Solving this non-homogenous recurrence relation.
This answer Solving this non-homogenous recurrence relation is too vague for me, and I don't know how to solve the equation using the method.
Best Answer
If we divide the equation $$a_{n+1}=(n+1)a_n\tag1$$ by $(n+1)!$ we get $$\frac{a_{n+1}}{(n+1)!}=\frac{a_n}{n!}.\tag2$$ Substituting $b_n=\frac{a_n}{n!}$ we get a homogeneous linear recurrence with constant coefficients, namely $$b_{n+1}=b_n.\tag3$$ The general solution of $(3)$ is $$b_n=c\tag4$$ where $c$ is an arbitrary constant. Hence the general solution of $(1)$ is $$a_n=cn!.\tag5$$