Explicit Formulas & Induction (Combinatorics) So my problem is:

Let $a_0 = 1$, and let $a_{n+1} = (10a_n) − 3$. Find an explicit formula for $a_n.$

I've come up with this sequence of the first few values:
1, 7, 67, 667, 6667, 66667, ...
However, I'm having trouble finding a formula, $a_m= $, to go along with this sequence that isn't the original. After figuring it out I would just need to use Induction and find the explicit formula but I'm stuck at this step. 
How do you come up with a formula for this sequence of values?  
 A: Just iterate the recursive relation:
\begin{align}a_{n} &= 10a_{n-1} - 3 \\
&= 10(10a_{n-2} - 3) - 3 \\
&= 10^2a_{n-2} - 33 \\
&= 10^2(10a_{n-3} - 3) - 33 \\
&= 10^3a_{n-3} - 333\\
&= \ldots\\
&= 10^{n-1}a_{1} - \underbrace{3\ldots3}_{n-1}\\
&= 10^{n-1}(10a_0 - 3) - \underbrace{3\ldots3}_{n-1}\\
&= 10^{n}a_{0} - \underbrace{3\ldots3}_{n}\\
&= 10^{n} - \underbrace{3\ldots3}_{n}\\
&= 10^{n} - 3\cdot\underbrace{1\ldots1}_{n}\\
&= 10^{n} - 3\sum_{i=0}^{n-1} 10^i\\
&= 10^{n} - 3\frac{10^n-1}{10 - 1}\\
&= 10^{n} - \frac{10^n-1}{3}\\
&= \frac{2 \cdot 10^n+1}{3}\\
\end{align}
Formally you would have to use induction, of course.
A: The explicit formula you are looking for is :
$$\tag{1}a_m=\dfrac{2 \times 10^m+1}{3}$$
Let us use induction to establish (1).
It is true for $a_0=1$. 
Let us assume (1) is valid for index $m$ ; let us show it is still valid for index $m+1$:
$$a_{m+1}=10\dfrac{2 \times 10^m+1}{3}-3=\dfrac{10(2 \times 10^m+1)}{3}-\dfrac{9}{3}=\dfrac{2 \times 10^{m+1}+10-9}{3}$$ giving
$$a_{m+1}=\dfrac{2 \times 10^{m+1}+1}{3}$$
which is the equivalent of (1) for index $m+1$.
