Need help with non-recursive definition So I'm trying to find a non-recursive definition for $b_n$.   I'm given $$b_0=1$$ $$b_{n+1}=2b_n-1$$
Does this mean I'm trying to find a number for $b_n$ that fits that algorithm?
Update:
Proof by induction.  Let P(n) be that for any $n$, $b_n=1$.
As our base case, we prove P(0), that $b_0=1$, which is obvious from the problem.
For our inductive step, assume for some n ∈ N that P(n) holds.  We prove that P(n+1) holds. 
$$b_{n+1}=2b_n-1$$
$$b_n=(b_{n+1}+1)/2$$
I'm not sure what to do next here.

Hint:
$$\begin{align*}
b_0 & = &1\\
b_1 & = 2b_0 -1 = (2\cdot 1)-1=&1\\
b_2 & = 2b_1 -1 = (2\cdot 1)-1=&1\\
 &\,\vdots& \vdots\,
\end{align*}$$
