Closed form for an exponential recurrence

Question

I was wondering what the closed form for a function $f(x)$ is when $f(x)=a*f(x-1)+b^\left(x-1\right)-c^\left(x+1\right)$, and $f(0) = d$ is given. Assume $a$, $b$, $c$, and $d$ are all greater than 1. It seems to be an exponential term based off of $a$, which makes sense, but I don't know how to make a closed form for the exact value.

Answer 1

$f(x)=a*f(x-1)+b^\left(x-1\right)-c^\left(x+1\right)$, and $f(0) = d$

The initial point of the recurrence is given only for $x=0$. But $f(x)$ is not given for $0<x<1$. This is not sufficient to determine $f(x)$ for non-integer $x$.

Since the wording of the problem is incomplete, the answer is impossible in generality. We can only answer to the question restricted to $x=n=$integer. $$f(n)=a\:f(n-1)+b^{n-1}-c^{n+1}\quad;\quad f(0) = d$$ $$af(n-1)=a\left(a\:f(n-2)+b^{n-2}-c^{n}\right)$$ $$…$$ $$a^kf(n-k)=a^k\left(a\:f(n-k-1)+b^{n-k-1}-c^{n-k+1}\right)$$ $$…$$ $$a^{n-2}f(2)=a^{n-2}\left(a\:f(1)+b^{1}-c^{3}\right)$$ $$a^{n-1}f(1)=a^{n-1}\left(a\:f(0)+b^{0}-c^{2}\right)$$ Summing all leads to : $$f(n)=a^nf(0)+\sum_{k=0}^{n-1} a^kb^{n-k-1} -\sum_{k=0}^{n-1} a^kc^{n-k+1}$$ $\sum_{k=0}^{n-1} a^kb^{n-k-1}=b^{n-1}\sum_{k=0}^{n-1} (\frac{a}{b})^k = b^{n-1}\frac{1-(\frac{a}{b})^n}{1-\frac{a}{b}}=\frac{b^n-a^n}{b-a}\quad$ : geometric series.

$\sum_{k=0}^{n-1} a^kc^{n-k+1}=c^2\frac{c^n-a^n}{c-a}$ $$f(n)=a^nd+\frac{b^n-a^n}{b-a}-c^2\frac{c^n-a^n}{c-a}$$