Induction: How to prove that $ab^n+cn+d$ is divisible by $m$. 
If $a+d$, $(b-1)c$, $ab-a+c$ are divisible by $m$, prove that
  $ab^n+cn+d$ is also divisible by $m$.

I want to prove this by induction. For proving $ab^{k+1}+c(k+1)+d$ is divisible by $m$, i want to prove that $ab^k(b-1)+c$ is divisible by $m$ and then add it to $ab^{k}+ck+d$. Any idea how to prove $ab^k(b-1)+c$ is divisible by $m$? Or is there a better way to solve the problem?
Thanks in advance.
 A: $\bmod m\!:\   \color{#0a0}{f_{\large n+1}\!-b f_{\large n}} =\, \overbrace{(1\!-\!b)c}^{\large \equiv\ 0}n \,\overbrace{-\color{#c00}d\,b+\color{#c00}d+c}^{\large \equiv\ 0\ \ {\rm by}\ \ \color{#c00}{d}\ \equiv\ -a}\!\equiv\color{#0a0} 0,\ $ so $\ f_n\equiv 0\,\Rightarrow\,\color{#0a0}{f_{n+1}\equiv bf_n\equiv} 0$
A: As I show here, such results boil down to the first $\,2\,$ terms of the Binomial Theorem, viz.
$\!\bmod m\!:\ c\equiv -a(b\!-\!1)\,$ so $\ \color{#0a0}0\equiv (b\!-\!1)c\equiv -\color{#0a0}{a(b\!-\!1)^2}$ so only $1$st $2$ terms survive below
$\qquad\quad\ \ \  a(1+b\!-\!1)^n =\, \color{#c00}{a\, +\,} n\,\underbrace{a(b\!-\!1)}_{\Large \equiv\ \color{#c00}{-c}}\ +\ \underbrace{\color{#0a0}{a(b\!-\!1)^2}(\cdots)}_{\large \equiv\ \color{#0a0}0}\ \,$ so adding $\ cn+d\ $ we get
$\quad\  \Rightarrow\ \ ab^n +cn+d\, \equiv\, \color{#c00}{a-c\,n}+cn\! +\! d \,\equiv\, a\!+\!d\,\equiv\, 0\ \ \ $ QED
Remark $ $ If you require an inductive proof then do so for the first $2$ terms in the Binomial Theorem. It's easy - the inductive step amounts to multiplying by $\,1\!+\!a\pmod{\!a^2},\,$ viz.
$\!\begin{align}{\rm mod}\,\ \color{#c00}{a^2}\!:\,\  (1+ a)^n\, \ \  \equiv&\,\ \ 1 + na\qquad\qquad\,\ \  {\rm i.e.}\ \ P(n)\\[1pt]
\Rightarrow\ \ (1+a)^{\color{}{n+1}}\! \equiv &\  (1+na)(1 + a)\quad\, {\rm by}\ \ 1+a \ \ \rm times\ prior\\ 
\equiv &\,\ \ 1+ na+a+n\color{#c00}{a^2}\\ 
 \equiv &\,\ \ 1\!+\! (n\!+\!1)a\qquad\ \ \ {\rm i.e.}\ \ P(\color{}{n\!+\!1})\\[2pt]  
  \end{align}$
A: We need to proceed by


*

*base case: $n=0 \implies ab^0+c0+d=a+d$

*induction step: assume $m|ab^n+cn+d$ then


$$ab^{n+1}+c(n+1)+d=ab^{n+1}+bcn+bd-bcn-bd+c(n+1)+d=$$
$$=b(ab^{n}+cn+b)-(b-1)cn-bd+c+d$$
then note 


*

*$m|a+d \implies m|ab+bd \implies m|bd+a-c\implies m|bd-c-d$
A: Without induction. Let's do some tagging 1st
$$m \mid a+d \tag{1}$$
$$m \mid (b-1)c \tag{2}$$
$$m \mid ab-a+c \tag{3}$$
then
$$m \mid ab^n+cn+d \iff m \mid a\left(b^n-1\right)+cn+\color{red}{a+d} \overset{(1)}{\iff}\\
m \mid a\left(b^n-1\right)+cn \iff \\
m \mid a(b-1)\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)+cn \iff \\
m \mid (ab-a+c-c)\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)+cn \iff \\
m \mid \color{red}{(ab-a+c)\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)}-c\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)+cn \overset{(3)}{\iff}\\
m \mid c\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)-cn \iff \\
m \mid c\left(b^{n-1}-1\right)+c\left(b^{n-2}-1\right)+...+c\left(b^2-1\right)+\color{red}{c\left(b-1\right)} \overset{(2)}{\iff}\\ 
m \mid c\left(b^{n-1}-1\right)+c\left(b^{n-2}-1\right)+...+c\left(b^2-1\right)$$
which is true because for $\forall k\geq 2$ we have
$$c\left(b^k-1\right)=\color{red}{c\left(b-1\right)}\left(b^{k-1}+b^{k-2}+...+b^{2}+b+1\right)$$
and from $(2)$
$$m \mid c\left(b^k-1\right)$$
