Assume we have an arithmetic sequence $f_n$ that has a non-recursive definition as follows:
$$f_n=2n+1$$
This sequence can also be defined recursively:
$$f_n=2+f_{n-1}$$
$$f_0=1$$
The same can be done if we assume f_n is a geometric sequence:
$$f_n=3^n$$
$$f_n=3f_{n-1}$$
$$f_0=1$$
Note that if either the non-recursive rule or the recursive definition of a geometric sequence or an arithmetic sequence is known, the other definition can be derived. This is possible, however, only when the sequence is known to be distinctly arithmetic or geometric. In other words, the recursive definition should be $f_n=a*f_{n-1}$ or $f_n=a+f_{n-1}$. My question is that is there a proof if it possible to derive a general rule for a sequence with the combined recursive definitions of an arithmetic and a geometric sequence which looks like:
$$f_n=af_{n-1}+b$$
$$f_0=c$$
If it is possible, what would be the way to derive the general rule ?
P.S. I am not that much informed in this area, and because of this, I would be very glad if you point out the unclear or wrong parts of the question. Constructive feedback is always useful 🙂
Best Answer
If $a=1$ we have an arithmetic progression with $f_n=bn+c$
Otherwise it would be easier if we could find a $k$ such that $g_n=f_n+k$ and $g_n=ag_{n-1}$ making $g_n=a^ng_0$
so we want $f_n+k = a(f_{n-1}+k)$ while we know $f_n = af_{n-1}+b$
and this means we need $k=\frac{b}{a-1}$ and we do not have division by zero as $a \not=1$, so $$f_n=g_n -\frac{b}{a-1} = a^n g_0 -\frac{b}{a-1}= a^n\left(c+\frac{b}{a-1}\right)-\frac{b}{a-1} $$