Formula for b ... b ( b ( a ) + c ) + c) + ... c 1) Let say we have a number A.  
2) We multiply it by B and add C to it. 
3) We repeat this action for N times.
For example for N = 3, A = 1, B = 3, C = 1, We have:
3 ( 3 ( 3 ( 1 ) + 1 ) + 1 ) + 1 ) = 40
What kind of progression do we have? Is it arithmetic or geometric? Is there any formula for the nth element or sum of the sequence until the nth element?
 A: Distributing all the multiplications, your expression simplifies to $B^NA + \sum_{k=0}^{N-1}B^kC$. You can prove this by induction on $N$. Except for the first term, this is a geometric series, which has closed form $$B^NA + \frac{C(1-B^N)}{(1-B)}.$$
In the simpler case that $C = A$, we really do get just a geometric series, which simplifies to $$\frac{A(1-B^{N+1})}{(1-B)}.$$
A: We want to look at the recurrence,
$$x_{n+1}=bx_{n}+c \tag{1}$$
With some initial value $x_0=a$. 
We may notice that there is a number $L$ such that,
$$L=bL+c \tag{2}$$
In the case $b \neq 1$, this number is $\frac{c}{1-b}$. We shall assume that, otherwise the solution is easy.
Subtracting the second equation from the first gives,
$$(x_{n+1}-L)=b(x_{n}-L)$$
So,
$$x_{n}-L=b^{n}(x_{0}-L)$$
(If your having trouble seeing this define $a_n=x_n-L$ then $a_{n+1}=ba_{n}$ etc) 
Which tells us that,
$$x_{n}=L+b^{n}(a-L)$$
Or more explicitly,
$$x_{n}=\frac{c}{1-b}+b^{n}(a-\frac{c}{1-b})$$
In the case $b=1$ obviously (and simple to prove),
$$x_{n}=x_{0}+cn=a+cn$$
